Difference between revisions of "2024 AMC 12A Problems/Problem 13"
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(graph by Technodoggo) | (graph by Technodoggo) | ||
− | ==Solution 3(Derivatives)== | + | ==Solution 3 (Derivatives)== |
− | + | Since y is a proper function of x, the axis of symmetry is vertical (and so answer choices A and B are impossible). | |
+ | Where a function crosses a vertical axis of symmetry, it must have either a vertical asymptote (clearly impossible for this given function, since linear functions and exponential function are total functions) or a local maximum or minimum (derivative 0). | ||
+ | |||
+ | So the derivative is equal to 0: | ||
<math>\frac{d[e^{x+1}+e^{-x}-2]}{dx} = 0 = e^{x+1}-e^{-x}</math> | <math>\frac{d[e^{x+1}+e^{-x}-2]}{dx} = 0 = e^{x+1}-e^{-x}</math> | ||
− | + | ||
+ | To solve for <math>x</math>, make a substitution: | ||
<math>u = e^x</math> | <math>u = e^x</math> | ||
<math>0 = e*u - \frac{1}{u}</math> -> | <math>0 = e*u - \frac{1}{u}</math> -> | ||
Line 63: | Line 67: | ||
<math>2x = -1</math> -> | <math>2x = -1</math> -> | ||
<math>x = -\frac{1}{2}</math> | <math>x = -\frac{1}{2}</math> | ||
− | |||
− | + | Reflect <math>\left(-1, \frac{1}{2}\right)</math> over <math>x = -\frac{1}{2}</math>, giving answer choice <math>\boxed{\textbf{(D) }\left(0,\dfrac12\right)}.</math> | |
==Solution 4 == | ==Solution 4 == | ||
Line 85: | Line 88: | ||
~KEVIN_LIU | ~KEVIN_LIU | ||
+ | |||
+ | ==Solution 5 (Graph)== | ||
+ | |||
+ | [[Image:2024_amc12A_p13.PNG|thumb|center|600px| ]] | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
==Solution 6 == | ==Solution 6 == | ||
− | <cmath> f(-x-1) = e^{-x-1+1} + e^{-(-x-1)} - 2 = e^{-x} + e^{x+1 | + | <cmath> f(-x-1) = e^{-x-1+1} + e^{-(-x-1)} - 2 = e^{-x} + e^{x+1} - 2 = f(x) </cmath> |
− | + | So the axis is <math>\frac{x + (-x + 1)}{2} = -1/2</math> | |
− | + | Point <math>(-1, 1/2)</math> reflects over the axis <math>x=-1/2</math> to the point <math>\boxed{\textbf{(D) }\left(0,\dfrac12\right)}</math> | |
+ | |||
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
==Solution 7 == | ==Solution 7 == | ||
− | + | Setting the axis of symmetry as <math>x=a</math> implies <math>f(2a-x) = f(x)</math> | |
+ | |||
+ | <cmath>\begin{align*} | ||
+ | f(2a-x) &= f(x) \ | ||
+ | e^{2a-x+1} + e^{-(2a-x)} - 2 &= e^{x+1} + e^{-x} - 2 | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Analyzing the terms <math>e^{2a-x+1}</math> and <math>e^{-x}</math> tells us that <math>2a+1 = 0</math> | ||
+ | |||
+ | This reveals that <math>a = -1/2</math> | ||
− | |||
− | |||
− | |||
− | |||
− | point (-1, 1/2) | + | Finally, point <math>(-1, 1/2)</math> reflected over the axis <math>x=-1/2</math> is the point <math>\boxed{\textbf{(D) }\left(0,\dfrac12\right)}</math> |
+ | |||
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
Latest revision as of 09:34, 24 November 2024
Contents
[hide]Problem
The graph of has an axis of symmetry. What is the reflection of the point over this axis?
Solution 1
The line of symmetry is probably of the form for some constant . A vertical line of symmetry at for a function exists if and only if ; we substitute and into our given function and see that we must have
for all real . Simplifying:
If , then for all real ; this is clearly impossible, so let . Thus, our line of symmetry is , and reflecting over this line gives
~Technodoggo
Solution 2 (Graphing cheese)
Consider the graphs of and . A rough sketch will show that they intercept somewhere between -1 and 0 and the axis of symmetry is vertical. Thus, is the only possible answer.
Note: You can more rigorously think about the solution by noting that since the derivative of the power that e is raised to in one equation is equal to the derivative of the power that e is raised to multiplied by and both equations are subtracted by 1, then the sum of both equations 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 , giving us the axis of symmetry.
~woeIsMe
(graph by Technodoggo)
Solution 3 (Derivatives)
Since y is a proper function of x, the axis of symmetry is vertical (and so answer choices A and B are impossible). Where a function crosses a vertical axis of symmetry, it must have either a vertical asymptote (clearly impossible for this given function, since linear functions and exponential function are total functions) or a local maximum or minimum (derivative 0).
So the derivative is equal to 0:
To solve for , make a substitution: -> -> -> -> -> -> ->
Reflect over , giving answer choice
Solution 4
so f(0) = f(-1) , f(1) =f(-2)
then f(x) is symmetric about x=-1/2
point (-1, 1/2) reflects over axis x=-1/2 is point ( 0, 1/2) answer choice D
~luckuso
Solution 5
Notice that this is simply a transformation of , specifically . We know that has a parabolic shape with an axis of symmetry . Since this is simple a transformation of to the left of the origin, we have the axis of symmetry is giving us
~KEVIN_LIU
Solution 5 (Graph)
Solution 6
So the axis is
Point reflects over the axis to the point
Solution 7
Setting the axis of symmetry as implies
Analyzing the terms and tells us that
This reveals that
Finally, point reflected over the axis is the point
See also
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.