Question 754·Hard·Math

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The function $f$ is defined for all real numbers $x$ by

f(x)=b^{x}+b^{-x},

where $b>1$ is a constant. Given that $f(1)=5$ , what is $f(3)$ ?

An exponential function containing both $b^x$ and $b^{-x}$ may give you one output and ask for another. Use the known output to find the allowed base with a Desmos regression, then evaluate the function at the requested input. Cubing a sum includes cross terms; it does not cube each term alone.

Hints

Hint 1
$f(1)$ means the output when $x=1$ . Replace $x$ in both exponents to turn the known output into an equation involving $b$ and its reciprocal, $1/b$ .
Hint 2
The equation may have a base above $1$ and another below it. A Desmos regression can find a base that fits, but you must keep the condition $b>1$ .
Hint 3
To find $f(3)$ , put $3$ in both exponents of the function rule. A negative exponent makes a reciprocal; it does not tell you to cube the earlier output.

Step-by-step

Approach 1: Find the base in Desmos

Step 1
Turn the known output into an equation
$f(1)=5$ says the function outputs $5$ when its input is $1$ . Put $1$ in both exponents:
$b^1+b^{-1}=5.$
Step 2
Rewrite the negative exponent
Rewrite the negative exponent as a reciprocal:
$b+\frac{1}{b}=5.$
Step 3
Find the allowed base
In Desmos, call the unknown base $x_1$ so it can be fitted. Type $x_1+\frac{1}{x_1}\sim5\{x_1>1\}$ ; $\sim$ tells Desmos to fit the equation, and the braces enforce $b>1$ . Under PARAMETERS, the regression gives $x_1\approx4.7913$ . That's the allowed value of $b$ ; without the restriction, Desmos could find the reciprocal value below $1$ .
Step 4
Evaluate the requested output
Add $f(x)=x_1^x+x_1^{-x}$ , using the fitted $x_1$ for $b$ . Then type $f(3)$ . Desmos prints $110$ . The output $f(3)$ adds $b^3$ and $b^{-3}$ ; it does not cube $f(1)$ . So $f(3)=110$ . Choice C.

Approach 2: See why the cube needs a correction

Step 1
Keep the extra terms when cubing
From $f(1)=5$ , you know $b+b^{-1}=5$ . The cube-of-a-sum identity is $(u+v)^3=u^3+v^3+3uv(u+v)$ ; its last part collects the cross terms. Apply it with $u=b$ and $v=b^{-1}$ :
$(b+b^{-1})^3=b^3+b^{-3}+3(b\cdot b^{-1})(b+b^{-1}).$
Step 2
Simplify the reciprocal product
A number times its reciprocal is $1$ , so simplify the cross-term factor:
$(b+b^{-1})^3=b^3+b^{-3}+3(b+b^{-1}).$
Step 3
Identify the function outputs
Name the sums using the function rule:
$(f(1))^3=f(3)+3f(1).$
Step 4
Use the output you know
Substitute the given $f(1)=5$ into the relation between the function outputs:
$5^3=f(3)+3(5).$
Step 5
Isolate the output you need
Subtract the cross terms $3(5)$ from both sides to leave $f(3)$ alone:
$f(3)=5^3-3(5).$
Step 6
Calculate the corrected cube
Type $5^3-3(5)$ in Desmos. It prints $110$ , the sum $b^3+b^{-3}$ . So $f(3)=110$ . Choice C.

Question 754·Hard·Math

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