Asked by Anonymous
How many critical points does the function f(x) = e^x - x^3 have?
Answers
Answered by
Steve
wherever e^x - 3x^2 = 0
That is, where e^x = 3x^2
You might not know just where they are, but you do know that 3x^2 is a parabola with vertex at (0,0).
Since e^x is kind of flat near there, you know that for x<0 there will be an intersection. Now the question is, will there also be one where x>0?
at x=0, e^x = 1 > 3x^2 = 0
at x=1, e^x = e < 3x^2 = 3
So, somewhere in the interval (0,1) there will be another intersection.
at x=4, e^4 ? 54 > 3x^2 = 27
So, in the interval (1,4) there will be another intersection. After that e^x remains greater than 3x^2.
So, e^x-x^3 has three critical points, as seen at
http://www.wolframalpha.com/input/?i=e%5Ex+-+x%5E3+for+-1%3Cx%3C4
That is, where e^x = 3x^2
You might not know just where they are, but you do know that 3x^2 is a parabola with vertex at (0,0).
Since e^x is kind of flat near there, you know that for x<0 there will be an intersection. Now the question is, will there also be one where x>0?
at x=0, e^x = 1 > 3x^2 = 0
at x=1, e^x = e < 3x^2 = 3
So, somewhere in the interval (0,1) there will be another intersection.
at x=4, e^4 ? 54 > 3x^2 = 27
So, in the interval (1,4) there will be another intersection. After that e^x remains greater than 3x^2.
So, e^x-x^3 has three critical points, as seen at
http://www.wolframalpha.com/input/?i=e%5Ex+-+x%5E3+for+-1%3Cx%3C4
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