2024 AMC 12A Problems/Problem 13

Problem

The graph of $y=e^{x+1}+e^{-x}-2$ has an axis of symmetry. What is the reflection of the point $(-1,\tfrac{1}{2})$ over this axis?

$\textbf{(A) }\left(-1,-\frac{3}{2}\right)\qquad\textbf{(B) }(-1,0)\qquad\textbf{(C) }\left(-1,\tfrac{1}{2}\right)\qquad\textbf{(D) }\left(0,\frac{1}{2}\right)\qquad\textbf{(E) }\left(3,\frac{1}{2}\right)$

Solution 1

The line of symmetry is probably of the form $x=a$ for some constant $a$ . A vertical line of symmetry at $x=a$ for a function $f$ exists if and only if $f(a-b)=f(a+b)$ ; we substitute $a-b$ and $a+b$ into our given function and see that we must have

$e^{a-b+1}+e^{-(a-b)}-2=e^{a+b+1}+e^{-(a+b)}-2$

for all real $b$ . Simplifying:

\begin{align*} e^{a-b+1}+e^{-(a-b)}-2&=e^{a+b+1}+e^{-(a+b)}-2 \\ e^{a-b+1}+e^{b-a}&=e^{a+b+1}+e^{-a-b} \\ e^{a-b+1}-e^{-a-b}&=e^{a+b+1}-e^{b-a} \\ e^{-b}\left(e^{a+1}-e^{-a}\right)&=e^b\left(e^{a+1}-e^{-a}\right). \\ \end{align*}

If $e^{a+1}-e^{-a}\neq0$ , then $e^{-b}=e^b$ for all real $b$ ; this is clearly impossible, so let $e^{a+1}-e^{-a}=0\implies a+1=-a\implies a=-\dfrac12$ . Thus, our line of symmetry is $x=-\dfrac12$ , and reflecting $\left(-1,\dfrac12\right)$ over this line gives $\boxed{\textbf{(D) }\left(0,\dfrac12\right)}.$

~Technodoggo

Solution 2 (Graphing)

Consider the graphs of $y=e^{x+1}-1$ and $y=e^{-x}-1$ . A rough sketch will show that they intercept somewhere between -1 and 0 and the axis of symmetry is vertical. Thus, $\boxed{\textbf{(D) }\left(0,\dfrac12\right)}.$ is the only possible answer.

Note: You can more rigorously think about the solution by noting that since that the absolute value of the derivative of the power that e is raised to is the same, and they are both subtracted by 1, then the sum of both functions will be the same from one side of the interception to the other. Setting both equations equal to each other, it is trivial to see $x=-1/2$ , giving us the axis of symmetry.

(someone insert the graph pls)

~woeIsMe

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2024 AMC 12A Problems/Problem 13

Contents

Problem

Solution 1

Solution 2 (Graphing)

See also