For a constant , define . If , what is the value of ?

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SAT Nonlinear Functions Question 27 | Free SAT Prep | Aniko

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Question 27·Hard·Nonlinear Functions

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For a constant $t>1$ , define $f(x)=t^x+t^{-x}$ . If $f(1)=7$ , what is the value of $f(3)$ ?

Your answer

(Express the answer as an integer)

For symmetric expressions like $t^x + t^{-x}$ , first use any given function value to find a simple base expression like $t + t^{-1}$ . Then look for algebraic identities (such as squaring or cubing that base expression) that produce the higher powers you need, and carefully isolate the desired term. On the SAT, it is faster to manipulate these identities symbolically than to solve for $t$ itself, and it helps to remember or quickly re-derive patterns like $(t + t^{-1})^3 = t^3 + t^{-3} + 3(t + t^{-1})$ .

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Use the given function value

First write $f(1)$ in terms of $t$ using the definition $f(x)=t^x+t^{-x}$ . What equation does $f(1)=7$ give you?

Connect $f(1)$ and $f(3)$

You know $t+t^{-1}$ . You need $t^3+t^{-3}$ . Think about how cubing $t+t^{-1}$ might produce $t^3$ and $t^{-3}$ as part of the expansion.

Expand a cube

Use $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ with $a=t$ and $b=t^{-1}$ . After expanding $(t+t^{-1})^3$ , group the terms to isolate $t^3+t^{-3}$ .

Substitute and simplify

Once you have an expression for $t^3+t^{-3}$ in terms of $t+t^{-1}$ , replace $t+t^{-1}$ with 7 and carefully do the arithmetic.

Desmos Guide

Represent the known sum

Type a = 7 in Desmos to stand for $t + 1/t = 7$ .

Express $t^3 + t^{-3}$ in terms of $a$

In a new line, type b = a^3 - 3a. This matches the identity $t^3 + t^{-3} = (t + t^{-1})^3 - 3(t + t^{-1})$ with $a = t + t^{-1}$ .

Read off the value of $f(3)$

Look at the numeric value that Desmos shows for b; that value is $f(3)$ .

Step-by-step Explanation

Use the definition of $f(x)$ to interpret $f(1)$

We are told that $f(x)=t^x+t^{-x}$ and that $f(1)=7$ .

f(1)=t^1+t^{-1}=t+t^{-1}=7.

This gives us a key relationship: $t+t^{-1}=7$ .

Write $f(3)$ in terms of $t$

We want $f(3)$ .

By the definition of $f$ ,

f(3)=t^3+t^{-3}.

So our goal is to express $t^3+t^{-3}$ in terms of $t+t^{-1}$ , which we already know equals 7.

Relate $t^3+t^{-3}$ to $t+t^{-1}$ using an algebraic identity

Consider the cube of the sum $t+t^{-1}$ :

(t+t^{-1})^3 = t^3 + 3t^2\cdot t^{-1} + 3t\cdot (t^{-1})^2 + t^{-3}.

Simplify the middle terms using exponent rules:

$t^2\cdot t^{-1}=t^{2-1}=t$
$t\cdot (t^{-1})^2 = t\cdot t^{-2}=t^{1-2}=t^{-1}$

(t+t^{-1})^3 = t^3 + 3t + 3t^{-1} + t^{-3}.

Group the symmetric terms:

(t+t^{-1})^3 = (t^3 + t^{-3}) + 3(t + t^{-1}).

Therefore,

t^3 + t^{-3} = (t+t^{-1})^3 - 3(t+t^{-1}).

Substitute the known value and compute $f(3)$

We know $t+t^{-1}=7$ , so substitute into the expression for $t^3+t^{-3}$ :

\begin{aligned} t^3 + t^{-3} &= (t+t^{-1})^3 - 3(t+t^{-1}) \\ &= 7^3 - 3\cdot 7. \end{aligned}

Compute:

$7^3 = 343$
$3\cdot 7 = 21$

t^3 + t^{-3} = 343 - 21 = 322.

Since $f(3)=t^3+t^{-3}$ , the value of $f(3)$ is 322.

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation