Practice · Plug special inputs into the identity · Sensim Math

Variant 1 easy

A function $f(x)$ satisfies $f(x)f(y)=f(x+y)+f(x-y)$ for all numbers $x$ and $y$ , and $f(1)=2$ . Find the value of $f(0)+f(2)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The identity holds for every pair $x$ , $y$ , so we may choose any pair we like. Pick pairs that turn the identity into an equation with a single unknown value of $f$ .

Givens

$f(x)f(y)=f(x+y)+f(x-y)$ for all $x$ , $y$ .
$f(1)=2$ .

Unknowns

The value of $f(0)+f(2)$ .

Constraints

Only $f(1)$ is known at the start.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #13 Convert to Algebra#7 Identify Subproblems

Rather than the general identity, solve easier special cases: $y=0$ makes $x+y$ and $x-y$ equal, and $x=y$ makes $x-y=0$ . With $x=1$ these give $f(0)$ first and then $f(2)$ .

3 · Execute3 carry out the plan

1Use x = 1, y = 0

#9 Solve an Easier Related Problem 8.F.A.1

Both

x+y

and

x-y

become

1

.

f(1)f(0)=f(1)+f(1)\Rightarrow 2\times f(0)=4 \Rightarrow f(0)=2

Choosing

y=0

collapses the right side to twice one known value.

2Use x = 1, y = 1

#13 Convert to Algebra 8.EE.C.7

Now

x+y=2

and

x-y=0

, and

f(0)

is already known.

f(1)f(1)=f(2)+f(0)\Rightarrow 4=f(2)+2 \Rightarrow f(2)=2

Each substitution brings in only one new unknown, so it can be solved.

3Combine

#7 Identify Subproblems 8.EE.C.7

Put the two values into the expression.

f(0)+f(2)=1\times2+1\times2=4

The two pieces were found separately and are now just added.

Answer:

4

4 · Reviewdoes it hold up?

Test another pair for consistency: $x=2$ , $y=0$ gives $f(2)f(0)=2f(2)$ , that is $2\times2=4=2\times2$ . It holds.

Another way: Use $x=0$ , $y=1$ instead of $x=1$ , $y=0$ : $f(0)f(1)=f(1)+f(-1)$ . That brings in $f(-1)$ , so pair it with $x=1$ , $y=0$ ; the choice $y=0$ is the quickest because it needs no new values.

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating $f(0)$, $f(1)$, $f(2)$ as the outputs for specific inputs.
8.EE.C.7 Solve linear equations in one variable — Solving the resulting one-variable linear equations.

💡Takeaway. When a rule holds for all inputs, feed it the friendliest ones, like 0 and 1, first.

Variant 2 easy

A function $f(x)$ satisfies $f(x)f(y)=f(x+y)+f(x-y)$ for all numbers $x$ and $y$ , and $f(1)=4$ . Find the value of $f(0)+f(2)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The identity holds for every pair $x$ , $y$ , so we may choose any pair we like. Pick pairs that turn the identity into an equation with a single unknown value of $f$ .

Givens

$f(x)f(y)=f(x+y)+f(x-y)$ for all $x$ , $y$ .
$f(1)=4$ .

Unknowns

The value of $f(0)+f(2)$ .

Constraints

Only $f(1)$ is known at the start.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #13 Convert to Algebra#7 Identify Subproblems

Rather than the general identity, solve easier special cases: $y=0$ makes $x+y$ and $x-y$ equal, and $x=y$ makes $x-y=0$ . With $x=1$ these give $f(0)$ first and then $f(2)$ .

3 · Execute3 carry out the plan

1Use x = 1, y = 0

#9 Solve an Easier Related Problem 8.F.A.1

Both

x+y

and

x-y

become

1

.

f(1)f(0)=f(1)+f(1)\Rightarrow 4\times f(0)=8 \Rightarrow f(0)=2

Choosing

y=0

collapses the right side to twice one known value.

2Use x = 1, y = 1

#13 Convert to Algebra 8.EE.C.7

Now

x+y=2

and

x-y=0

, and

f(0)

is already known.

f(1)f(1)=f(2)+f(0)\Rightarrow 16=f(2)+2 \Rightarrow f(2)=14

Each substitution brings in only one new unknown, so it can be solved.

3Combine

#7 Identify Subproblems 8.EE.C.7

Put the two values into the expression.

f(0)+f(2)=1\times2+1\times14=16

The two pieces were found separately and are now just added.

Answer:

16

4 · Reviewdoes it hold up?

Test another pair for consistency: $x=2$ , $y=0$ gives $f(2)f(0)=2f(2)$ , that is $14\times2=28=2\times14$ . It holds.

Another way: Use $x=0$ , $y=1$ instead of $x=1$ , $y=0$ : $f(0)f(1)=f(1)+f(-1)$ . That brings in $f(-1)$ , so pair it with $x=1$ , $y=0$ ; the choice $y=0$ is the quickest because it needs no new values.

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating $f(0)$, $f(1)$, $f(2)$ as the outputs for specific inputs.
8.EE.C.7 Solve linear equations in one variable — Solving the resulting one-variable linear equations.

💡Takeaway. When a rule holds for all inputs, feed it the friendliest ones, like 0 and 1, first.

Variant 3 easy

A function $f(x)$ satisfies $f(x)f(y)=f(x+y)+f(x-y)$ for all numbers $x$ and $y$ , and $f(1)=1$ . Find the value of $2f(0)+f(2)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The identity holds for every pair $x$ , $y$ , so we may choose any pair we like. Pick pairs that turn the identity into an equation with a single unknown value of $f$ .

Givens

$f(x)f(y)=f(x+y)+f(x-y)$ for all $x$ , $y$ .
$f(1)=1$ .

Unknowns

The value of $2f(0)+f(2)$ .

Constraints

Only $f(1)$ is known at the start.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #13 Convert to Algebra#7 Identify Subproblems

Rather than the general identity, solve easier special cases: $y=0$ makes $x+y$ and $x-y$ equal, and $x=y$ makes $x-y=0$ . With $x=1$ these give $f(0)$ first and then $f(2)$ .

3 · Execute3 carry out the plan

1Use x = 1, y = 0

#9 Solve an Easier Related Problem 8.F.A.1

Both

x+y

and

x-y

become

1

.

f(1)f(0)=f(1)+f(1)\Rightarrow 1\times f(0)=2 \Rightarrow f(0)=2

Choosing

y=0

collapses the right side to twice one known value.

2Use x = 1, y = 1

#13 Convert to Algebra 8.EE.C.7

Now

x+y=2

and

x-y=0

, and

f(0)

is already known.

f(1)f(1)=f(2)+f(0)\Rightarrow 1=f(2)+2 \Rightarrow f(2)=-1

Each substitution brings in only one new unknown, so it can be solved.

3Combine

#7 Identify Subproblems 8.EE.C.7

Put the two values into the expression.

2f(0)+f(2)=2\times2+1\times(-1)=3

The two pieces were found separately and are now just added.

Answer:

3

4 · Reviewdoes it hold up?

Test another pair for consistency: $x=2$ , $y=0$ gives $f(2)f(0)=2f(2)$ , that is $(-1)\times2=-2=2\times(-1)$ . It holds.

Another way: Use $x=0$ , $y=1$ instead of $x=1$ , $y=0$ : $f(0)f(1)=f(1)+f(-1)$ . That brings in $f(-1)$ , so pair it with $x=1$ , $y=0$ ; the choice $y=0$ is the quickest because it needs no new values.

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating $f(0)$, $f(1)$, $f(2)$ as the outputs for specific inputs.
8.EE.C.7 Solve linear equations in one variable — Solving the resulting one-variable linear equations.

💡Takeaway. When a rule holds for all inputs, feed it the friendliest ones, like 0 and 1, first.

Variant 4 medium

A function $f(x)$ satisfies $f(x)f(y)=f(x+y)+f(x-y)$ for all numbers $x$ and $y$ , and $f(1)=3$ . Find the value of $f(0)+3f(2)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The identity holds for every pair $x$ , $y$ , so we may choose any pair we like. Pick pairs that turn the identity into an equation with a single unknown value of $f$ .

Givens

$f(x)f(y)=f(x+y)+f(x-y)$ for all $x$ , $y$ .
$f(1)=3$ .

Unknowns

The value of $f(0)+3f(2)$ .

Constraints

Only $f(1)$ is known at the start.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #13 Convert to Algebra#7 Identify Subproblems

Rather than the general identity, solve easier special cases: $y=0$ makes $x+y$ and $x-y$ equal, and $x=y$ makes $x-y=0$ . With $x=1$ these give $f(0)$ first and then $f(2)$ .

3 · Execute3 carry out the plan

1Use x = 1, y = 0

#9 Solve an Easier Related Problem 8.F.A.1

Both

x+y

and

x-y

become

1

.

f(1)f(0)=f(1)+f(1)\Rightarrow 3\times f(0)=6 \Rightarrow f(0)=2

Choosing

y=0

collapses the right side to twice one known value.

2Use x = 1, y = 1

#13 Convert to Algebra 8.EE.C.7

Now

x+y=2

and

x-y=0

, and

f(0)

is already known.

f(1)f(1)=f(2)+f(0)\Rightarrow 9=f(2)+2 \Rightarrow f(2)=7

Each substitution brings in only one new unknown, so it can be solved.

3Combine

#7 Identify Subproblems 8.EE.C.7

Put the two values into the expression.

f(0)+3f(2)=1\times2+3\times7=23

The two pieces were found separately and are now just added.

Answer:

23

4 · Reviewdoes it hold up?

Test another pair for consistency: $x=2$ , $y=0$ gives $f(2)f(0)=2f(2)$ , that is $7\times2=14=2\times7$ . It holds.

Another way: Use $x=0$ , $y=1$ instead of $x=1$ , $y=0$ : $f(0)f(1)=f(1)+f(-1)$ . That brings in $f(-1)$ , so pair it with $x=1$ , $y=0$ ; the choice $y=0$ is the quickest because it needs no new values.

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating $f(0)$, $f(1)$, $f(2)$ as the outputs for specific inputs.
8.EE.C.7 Solve linear equations in one variable — Solving the resulting one-variable linear equations.

💡Takeaway. When a rule holds for all inputs, feed it the friendliest ones, like 0 and 1, first.

Variant 5 easy

A function $f(x)$ satisfies $f(x)f(y)=f(x+y)+f(x-y)$ for all numbers $x$ and $y$ , and $f(1)=3$ . Find the value of $2f(0)+f(2)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The identity holds for every pair $x$ , $y$ , so we may choose any pair we like. Pick pairs that turn the identity into an equation with a single unknown value of $f$ .

Givens

$f(x)f(y)=f(x+y)+f(x-y)$ for all $x$ , $y$ .
$f(1)=3$ .

Unknowns

The value of $2f(0)+f(2)$ .

Constraints

Only $f(1)$ is known at the start.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #13 Convert to Algebra#7 Identify Subproblems

Rather than the general identity, solve easier special cases: $y=0$ makes $x+y$ and $x-y$ equal, and $x=y$ makes $x-y=0$ . With $x=1$ these give $f(0)$ first and then $f(2)$ .

3 · Execute3 carry out the plan

1Use x = 1, y = 0

#9 Solve an Easier Related Problem 8.F.A.1

Both

x+y

and

x-y

become

1

.

f(1)f(0)=f(1)+f(1)\Rightarrow 3\times f(0)=6 \Rightarrow f(0)=2

Choosing

y=0

collapses the right side to twice one known value.

2Use x = 1, y = 1

#13 Convert to Algebra 8.EE.C.7

Now

x+y=2

and

x-y=0

, and

f(0)

is already known.

f(1)f(1)=f(2)+f(0)\Rightarrow 9=f(2)+2 \Rightarrow f(2)=7

Each substitution brings in only one new unknown, so it can be solved.

3Combine

#7 Identify Subproblems 8.EE.C.7

Put the two values into the expression.

2f(0)+f(2)=2\times2+1\times7=11

The two pieces were found separately and are now just added.

Answer:

11

4 · Reviewdoes it hold up?

Test another pair for consistency: $x=2$ , $y=0$ gives $f(2)f(0)=2f(2)$ , that is $7\times2=14=2\times7$ . It holds.

Another way: Use $x=0$ , $y=1$ instead of $x=1$ , $y=0$ : $f(0)f(1)=f(1)+f(-1)$ . That brings in $f(-1)$ , so pair it with $x=1$ , $y=0$ ; the choice $y=0$ is the quickest because it needs no new values.

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating $f(0)$, $f(1)$, $f(2)$ as the outputs for specific inputs.
8.EE.C.7 Solve linear equations in one variable — Solving the resulting one-variable linear equations.

💡Takeaway. When a rule holds for all inputs, feed it the friendliest ones, like 0 and 1, first.

Variant 6 medium

A function $f(x)$ satisfies $f(x)f(y)=f(x+y)+f(x-y)$ for all numbers $x$ and $y$ , and $f(1)=1$ . Find the value of $f(0)+4f(2)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The identity holds for every pair $x$ , $y$ , so we may choose any pair we like. Pick pairs that turn the identity into an equation with a single unknown value of $f$ .

Givens

$f(x)f(y)=f(x+y)+f(x-y)$ for all $x$ , $y$ .
$f(1)=1$ .

Unknowns

The value of $f(0)+4f(2)$ .

Constraints

Only $f(1)$ is known at the start.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #13 Convert to Algebra#7 Identify Subproblems

Rather than the general identity, solve easier special cases: $y=0$ makes $x+y$ and $x-y$ equal, and $x=y$ makes $x-y=0$ . With $x=1$ these give $f(0)$ first and then $f(2)$ .

3 · Execute3 carry out the plan

1Use x = 1, y = 0

#9 Solve an Easier Related Problem 8.F.A.1

Both

x+y

and

x-y

become

1

.

f(1)f(0)=f(1)+f(1)\Rightarrow 1\times f(0)=2 \Rightarrow f(0)=2

Choosing

y=0

collapses the right side to twice one known value.

2Use x = 1, y = 1

#13 Convert to Algebra 8.EE.C.7

Now

x+y=2

and

x-y=0

, and

f(0)

is already known.

f(1)f(1)=f(2)+f(0)\Rightarrow 1=f(2)+2 \Rightarrow f(2)=-1

Each substitution brings in only one new unknown, so it can be solved.

3Combine

#7 Identify Subproblems 8.EE.C.7

Put the two values into the expression.

f(0)+4f(2)=1\times2+4\times(-1)=-2

The two pieces were found separately and are now just added.

Answer:

-2

4 · Reviewdoes it hold up?

Test another pair for consistency: $x=2$ , $y=0$ gives $f(2)f(0)=2f(2)$ , that is $(-1)\times2=-2=2\times(-1)$ . It holds.

Another way: Use $x=0$ , $y=1$ instead of $x=1$ , $y=0$ : $f(0)f(1)=f(1)+f(-1)$ . That brings in $f(-1)$ , so pair it with $x=1$ , $y=0$ ; the choice $y=0$ is the quickest because it needs no new values.

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating $f(0)$, $f(1)$, $f(2)$ as the outputs for specific inputs.
8.EE.C.7 Solve linear equations in one variable — Solving the resulting one-variable linear equations.

💡Takeaway. When a rule holds for all inputs, feed it the friendliest ones, like 0 and 1, first.

Variant 7 medium

A function $f(x)$ satisfies $f(x)f(y)=f(x+y)+f(x-y)$ for all numbers $x$ and $y$ , and $f(1)=4$ . Find the value of $3f(0)+f(2)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The identity holds for every pair $x$ , $y$ , so we may choose any pair we like. Pick pairs that turn the identity into an equation with a single unknown value of $f$ .

Givens

$f(x)f(y)=f(x+y)+f(x-y)$ for all $x$ , $y$ .
$f(1)=4$ .

Unknowns

The value of $3f(0)+f(2)$ .

Constraints

Only $f(1)$ is known at the start.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #13 Convert to Algebra#7 Identify Subproblems

Rather than the general identity, solve easier special cases: $y=0$ makes $x+y$ and $x-y$ equal, and $x=y$ makes $x-y=0$ . With $x=1$ these give $f(0)$ first and then $f(2)$ .

3 · Execute3 carry out the plan

1Use x = 1, y = 0

#9 Solve an Easier Related Problem 8.F.A.1

Both

x+y

and

x-y

become

1

.

f(1)f(0)=f(1)+f(1)\Rightarrow 4\times f(0)=8 \Rightarrow f(0)=2

Choosing

y=0

collapses the right side to twice one known value.

2Use x = 1, y = 1

#13 Convert to Algebra 8.EE.C.7

Now

x+y=2

and

x-y=0

, and

f(0)

is already known.

f(1)f(1)=f(2)+f(0)\Rightarrow 16=f(2)+2 \Rightarrow f(2)=14

Each substitution brings in only one new unknown, so it can be solved.

3Combine

#7 Identify Subproblems 8.EE.C.7

Put the two values into the expression.

3f(0)+f(2)=3\times2+1\times14=20

The two pieces were found separately and are now just added.

Answer:

20

4 · Reviewdoes it hold up?

Test another pair for consistency: $x=2$ , $y=0$ gives $f(2)f(0)=2f(2)$ , that is $14\times2=28=2\times14$ . It holds.

Another way: Use $x=0$ , $y=1$ instead of $x=1$ , $y=0$ : $f(0)f(1)=f(1)+f(-1)$ . That brings in $f(-1)$ , so pair it with $x=1$ , $y=0$ ; the choice $y=0$ is the quickest because it needs no new values.

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating $f(0)$, $f(1)$, $f(2)$ as the outputs for specific inputs.
8.EE.C.7 Solve linear equations in one variable — Solving the resulting one-variable linear equations.

💡Takeaway. When a rule holds for all inputs, feed it the friendliest ones, like 0 and 1, first.

Variant 8 medium

A function $f(x)$ satisfies $f(x)f(y)=f(x+y)+f(x-y)$ for all numbers $x$ and $y$ , and $f(1)=3$ . Find the value of $4f(0)+f(2)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The identity holds for every pair $x$ , $y$ , so we may choose any pair we like. Pick pairs that turn the identity into an equation with a single unknown value of $f$ .

Givens

$f(x)f(y)=f(x+y)+f(x-y)$ for all $x$ , $y$ .
$f(1)=3$ .

Unknowns

The value of $4f(0)+f(2)$ .

Constraints

Only $f(1)$ is known at the start.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #13 Convert to Algebra#7 Identify Subproblems

Rather than the general identity, solve easier special cases: $y=0$ makes $x+y$ and $x-y$ equal, and $x=y$ makes $x-y=0$ . With $x=1$ these give $f(0)$ first and then $f(2)$ .

3 · Execute3 carry out the plan

1Use x = 1, y = 0

#9 Solve an Easier Related Problem 8.F.A.1

Both

x+y

and

x-y

become

1

.

f(1)f(0)=f(1)+f(1)\Rightarrow 3\times f(0)=6 \Rightarrow f(0)=2

Choosing

y=0

collapses the right side to twice one known value.

2Use x = 1, y = 1

#13 Convert to Algebra 8.EE.C.7

Now

x+y=2

and

x-y=0

, and

f(0)

is already known.

f(1)f(1)=f(2)+f(0)\Rightarrow 9=f(2)+2 \Rightarrow f(2)=7

Each substitution brings in only one new unknown, so it can be solved.

3Combine

#7 Identify Subproblems 8.EE.C.7

Put the two values into the expression.

4f(0)+f(2)=4\times2+1\times7=15

The two pieces were found separately and are now just added.

Answer:

15

4 · Reviewdoes it hold up?

Test another pair for consistency: $x=2$ , $y=0$ gives $f(2)f(0)=2f(2)$ , that is $7\times2=14=2\times7$ . It holds.

Another way: Use $x=0$ , $y=1$ instead of $x=1$ , $y=0$ : $f(0)f(1)=f(1)+f(-1)$ . That brings in $f(-1)$ , so pair it with $x=1$ , $y=0$ ; the choice $y=0$ is the quickest because it needs no new values.

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating $f(0)$, $f(1)$, $f(2)$ as the outputs for specific inputs.
8.EE.C.7 Solve linear equations in one variable — Solving the resulting one-variable linear equations.

💡Takeaway. When a rule holds for all inputs, feed it the friendliest ones, like 0 and 1, first.

Variant 9 hard

A function $f(x)$ satisfies $f(x)f(y)=f(x+y)+f(x-y)$ for all numbers $x$ and $y$ , and $f(1)=5$ . Find the value of $f(0)+2f(2)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The identity holds for every pair $x$ , $y$ , so we may choose any pair we like. Pick pairs that turn the identity into an equation with a single unknown value of $f$ .

Givens

$f(x)f(y)=f(x+y)+f(x-y)$ for all $x$ , $y$ .
$f(1)=5$ .

Unknowns

The value of $f(0)+2f(2)$ .

Constraints

Only $f(1)$ is known at the start.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #13 Convert to Algebra#7 Identify Subproblems

Rather than the general identity, solve easier special cases: $y=0$ makes $x+y$ and $x-y$ equal, and $x=y$ makes $x-y=0$ . With $x=1$ these give $f(0)$ first and then $f(2)$ .

3 · Execute3 carry out the plan

1Use x = 1, y = 0

#9 Solve an Easier Related Problem 8.F.A.1

Both

x+y

and

x-y

become

1

.

f(1)f(0)=f(1)+f(1)\Rightarrow 5\times f(0)=10 \Rightarrow f(0)=2

Choosing

y=0

collapses the right side to twice one known value.

2Use x = 1, y = 1

#13 Convert to Algebra 8.EE.C.7

Now

x+y=2

and

x-y=0

, and

f(0)

is already known.

f(1)f(1)=f(2)+f(0)\Rightarrow 25=f(2)+2 \Rightarrow f(2)=23

Each substitution brings in only one new unknown, so it can be solved.

3Combine

#7 Identify Subproblems 8.EE.C.7

Put the two values into the expression.

f(0)+2f(2)=1\times2+2\times23=48

The two pieces were found separately and are now just added.

Answer:

48

4 · Reviewdoes it hold up?

Test another pair for consistency: $x=2$ , $y=0$ gives $f(2)f(0)=2f(2)$ , that is $23\times2=46=2\times23$ . It holds.

Another way: Use $x=0$ , $y=1$ instead of $x=1$ , $y=0$ : $f(0)f(1)=f(1)+f(-1)$ . That brings in $f(-1)$ , so pair it with $x=1$ , $y=0$ ; the choice $y=0$ is the quickest because it needs no new values.

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating $f(0)$, $f(1)$, $f(2)$ as the outputs for specific inputs.
8.EE.C.7 Solve linear equations in one variable — Solving the resulting one-variable linear equations.

💡Takeaway. When a rule holds for all inputs, feed it the friendliest ones, like 0 and 1, first.

Variant 10 hard

A function $f(x)$ satisfies $f(x)f(y)=f(x+y)+f(x-y)$ for all numbers $x$ and $y$ , and $f(1)=2$ . Find the value of $2f(0)+3f(2)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The identity holds for every pair $x$ , $y$ , so we may choose any pair we like. Pick pairs that turn the identity into an equation with a single unknown value of $f$ .

Givens

$f(x)f(y)=f(x+y)+f(x-y)$ for all $x$ , $y$ .
$f(1)=2$ .

Unknowns

The value of $2f(0)+3f(2)$ .

Constraints

Only $f(1)$ is known at the start.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #13 Convert to Algebra#7 Identify Subproblems

Rather than the general identity, solve easier special cases: $y=0$ makes $x+y$ and $x-y$ equal, and $x=y$ makes $x-y=0$ . With $x=1$ these give $f(0)$ first and then $f(2)$ .

3 · Execute3 carry out the plan

1Use x = 1, y = 0

#9 Solve an Easier Related Problem 8.F.A.1

Both

x+y

and

x-y

become

1

.

f(1)f(0)=f(1)+f(1)\Rightarrow 2\times f(0)=4 \Rightarrow f(0)=2

Choosing

y=0

collapses the right side to twice one known value.

2Use x = 1, y = 1

#13 Convert to Algebra 8.EE.C.7

Now

x+y=2

and

x-y=0

, and

f(0)

is already known.

f(1)f(1)=f(2)+f(0)\Rightarrow 4=f(2)+2 \Rightarrow f(2)=2

Each substitution brings in only one new unknown, so it can be solved.

3Combine

#7 Identify Subproblems 8.EE.C.7

Put the two values into the expression.

2f(0)+3f(2)=2\times2+3\times2=10

The two pieces were found separately and are now just added.

Answer:

10

4 · Reviewdoes it hold up?

Test another pair for consistency: $x=2$ , $y=0$ gives $f(2)f(0)=2f(2)$ , that is $2\times2=4=2\times2$ . It holds.

Another way: Use $x=0$ , $y=1$ instead of $x=1$ , $y=0$ : $f(0)f(1)=f(1)+f(-1)$ . That brings in $f(-1)$ , so pair it with $x=1$ , $y=0$ ; the choice $y=0$ is the quickest because it needs no new values.

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating $f(0)$, $f(1)$, $f(2)$ as the outputs for specific inputs.
8.EE.C.7 Solve linear equations in one variable — Solving the resulting one-variable linear equations.

💡Takeaway. When a rule holds for all inputs, feed it the friendliest ones, like 0 and 1, first.

Variant 11 hard

A function $f(x)$ satisfies $f(x)f(y)=f(x+y)+f(x-y)$ for all numbers $x$ and $y$ , and $f(1)=1$ . Find the value of $3f(0)+2f(2)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The identity holds for every pair $x$ , $y$ , so we may choose any pair we like. Pick pairs that turn the identity into an equation with a single unknown value of $f$ .

Givens

$f(x)f(y)=f(x+y)+f(x-y)$ for all $x$ , $y$ .
$f(1)=1$ .

Unknowns

The value of $3f(0)+2f(2)$ .

Constraints

Only $f(1)$ is known at the start.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #13 Convert to Algebra#7 Identify Subproblems

Rather than the general identity, solve easier special cases: $y=0$ makes $x+y$ and $x-y$ equal, and $x=y$ makes $x-y=0$ . With $x=1$ these give $f(0)$ first and then $f(2)$ .

3 · Execute3 carry out the plan

1Use x = 1, y = 0

#9 Solve an Easier Related Problem 8.F.A.1

Both

x+y

and

x-y

become

1

.

f(1)f(0)=f(1)+f(1)\Rightarrow 1\times f(0)=2 \Rightarrow f(0)=2

Choosing

y=0

collapses the right side to twice one known value.

2Use x = 1, y = 1

#13 Convert to Algebra 8.EE.C.7

Now

x+y=2

and

x-y=0

, and

f(0)

is already known.

f(1)f(1)=f(2)+f(0)\Rightarrow 1=f(2)+2 \Rightarrow f(2)=-1

Each substitution brings in only one new unknown, so it can be solved.

3Combine

#7 Identify Subproblems 8.EE.C.7

Put the two values into the expression.

3f(0)+2f(2)=3\times2+2\times(-1)=4

The two pieces were found separately and are now just added.

Answer:

4

4 · Reviewdoes it hold up?

Test another pair for consistency: $x=2$ , $y=0$ gives $f(2)f(0)=2f(2)$ , that is $(-1)\times2=-2=2\times(-1)$ . It holds.

Another way: Use $x=0$ , $y=1$ instead of $x=1$ , $y=0$ : $f(0)f(1)=f(1)+f(-1)$ . That brings in $f(-1)$ , so pair it with $x=1$ , $y=0$ ; the choice $y=0$ is the quickest because it needs no new values.

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating $f(0)$, $f(1)$, $f(2)$ as the outputs for specific inputs.
8.EE.C.7 Solve linear equations in one variable — Solving the resulting one-variable linear equations.

💡Takeaway. When a rule holds for all inputs, feed it the friendliest ones, like 0 and 1, first.

Variant 12 hard

A function $f(x)$ satisfies $f(x)f(y)=f(x+y)+f(x-y)$ for all numbers $x$ and $y$ , and $f(1)=2$ . Find the value of $3f(0)+2f(2)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

The identity holds for every pair $x$ , $y$ , so we may choose any pair we like. Pick pairs that turn the identity into an equation with a single unknown value of $f$ .

Givens

$f(x)f(y)=f(x+y)+f(x-y)$ for all $x$ , $y$ .
$f(1)=2$ .

Unknowns

The value of $3f(0)+2f(2)$ .

Constraints

Only $f(1)$ is known at the start.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #13 Convert to Algebra#7 Identify Subproblems

Rather than the general identity, solve easier special cases: $y=0$ makes $x+y$ and $x-y$ equal, and $x=y$ makes $x-y=0$ . With $x=1$ these give $f(0)$ first and then $f(2)$ .

3 · Execute3 carry out the plan

1Use x = 1, y = 0

#9 Solve an Easier Related Problem 8.F.A.1

Both

x+y

and

x-y

become

1

.

f(1)f(0)=f(1)+f(1)\Rightarrow 2\times f(0)=4 \Rightarrow f(0)=2

Choosing

y=0

collapses the right side to twice one known value.

2Use x = 1, y = 1

#13 Convert to Algebra 8.EE.C.7

Now

x+y=2

and

x-y=0

, and

f(0)

is already known.

f(1)f(1)=f(2)+f(0)\Rightarrow 4=f(2)+2 \Rightarrow f(2)=2

Each substitution brings in only one new unknown, so it can be solved.

3Combine

#7 Identify Subproblems 8.EE.C.7

Put the two values into the expression.

3f(0)+2f(2)=3\times2+2\times2=10

The two pieces were found separately and are now just added.

Answer:

10

4 · Reviewdoes it hold up?

Test another pair for consistency: $x=2$ , $y=0$ gives $f(2)f(0)=2f(2)$ , that is $2\times2=4=2\times2$ . It holds.

Another way: Use $x=0$ , $y=1$ instead of $x=1$ , $y=0$ : $f(0)f(1)=f(1)+f(-1)$ . That brings in $f(-1)$ , so pair it with $x=1$ , $y=0$ ; the choice $y=0$ is the quickest because it needs no new values.

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating $f(0)$, $f(1)$, $f(2)$ as the outputs for specific inputs.
8.EE.C.7 Solve linear equations in one variable — Solving the resulting one-variable linear equations.

💡Takeaway. When a rule holds for all inputs, feed it the friendliest ones, like 0 and 1, first.

Plug special inputs into the identity · 12 practice problems

Generated variants — 12

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