Question 742·Hard·Math

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Let the function $f$ be defined by

f(x) = a^{x-1} + a^{1-x}

where $a$ is a positive constant with $a \ne 1$ . If $f(3) = 5$ , what is the value of $f(5)$ ?

The exponents $x-1$ and $1-x$ are opposites, so the function contains reciprocal powers. When one output is given and another is asked for, you can use a Desmos regression to find a base, then evaluate the function. There are two possible bases, but they swap the terms and give the same output.

Hints

Hint 1
The input in $f(3)$ is $3$ . Put it into both exponents in the rule. What equation do you get when you set the resulting expression equal to the given output?
Hint 2
A negative exponent means a reciprocal: $a^{-2}=1/a^2$ . Replacing $a$ with $1/a$ switches the two terms without changing their sum. Why is it enough to fit the base greater than $1$ ?
Hint 3
Once the regression finds $a$ , define the original function in Desmos and enter $f(5)$ . Using the same rule keeps both exponents attached to the input you’re evaluating.

Step-by-step

Approach 1: Fit a base, then evaluate in Desmos

Step 1
Turn the given output into an equation
$f(3)$ means put $3$ in for $x$ . The given output is $5$ , so replace both copies of $x$ and set the result equal to $5$ :
$a^{3-1}+a^{1-3}=5.$
Step 2
Simplify the exponents
Simplify $3-1$ and $1-3$ to get the equation for the base:
$a^2+a^{-2}=5.$
Step 3
Fit one of the possible bases
Since $a>0$ and $a\ne1$ , either $a$ or $1/a$ is greater than $1$ . Replacing $a$ with $1/a$ swaps the two terms of $f(x)$ , so either base gives the same output. Type $a^2+a^{-2}\sim5\left\{a>1\right\}$ ; the $\sim$ tells Desmos to fit $a$ rather than make a slider. Under PARAMETERS, it shows $a\approx2.1889$ .
Step 4
Evaluate the requested output
With the fitted $a$ in Desmos, type the original rule $f(x)=a^{x-1}+a^{1-x}$ , then type $f(5)$ . Desmos displays $23$ . So the value of $f(5)$ is $23$ . Choice C.

Approach 2: Use the square of the known sum

Step 1
Square the sum from the given output
Instead of finding $a$ , start with $a^2+a^{-2}=5$ . Squaring produces the fourth powers needed for the new output. Square both sides:
$(a^2+a^{-2})^2=5^2.$
Step 2
Expand the square
Expand the left side:
$a^4+2a^2a^{-2}+a^{-4}=5^2.$
Squaring a sum creates two cross terms as well as the two squared terms.
Step 3
Simplify the middle product
A negative exponent makes a reciprocal, so $a^2a^{-2}=1$ because $a>0$ . Replace that product with $1$ :
$a^4+2+a^{-4}=5^2.$
Step 4
Identify the expression you need
Put $5$ in for $x$ in the rule. This gives the two outer terms of the squared sum:
$f(5)=a^4+a^{-4}.$
Step 5
Isolate those terms
Subtract $2$ from both sides to leave the expression for $f(5)$ :
$a^4+a^{-4}=5^2-2.$
Step 6
Calculate the value
Type $5^2-2$ in Desmos. It prints $23$ , so $f(5)=23$ . Choice C.

Question 742·Hard·Math

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

Lessons that teach this