Is #f(x)=1-x-e^(x)/x^2# concave or convex at #x=-1#?

1 Answer
Noah G
Aug 8, 2017

Convex

Explanation:

The second derivative is determinative of concavity. Start by finding the first derivative.

#f'(x) = -1 - (e^x(x^2) - e^x(2x))/(x^2)^2#

#f'(x) = 1 - (x(xe^x - 2))/x^4#

#f'(x) = 1 - (xe^x - 2)/x^3#

Now differentiate this.

#f''(x) = 0 - ((e^x + xe^x)x^3 - (xe^x - 2)(3x^2))/(x^3)^2#

#f''(x) = - (x^3e^x + x^4e^x - 3x^3e^x + 6x^2)/x^6#

#f''(x) = -(x^4e^x - 2x^3e^x + 6x^2)/x^6#

We now evaluate the point within the second derivative.

#f''(-1) = -((-1)^4e^-1 - 2(-1)^3e^-1 + 6(-1)^2)/(-1)^6#

#f''(-1) = -(1/e -2(-1)/e + 6)/1#

#f''(-1) =- (1/e + 2/e + 6)/1#

This is obviously negative, therefore, the function is convex at #x = -1#. If the second derivative had had a positive value at #x = -1#, then we would have dubbed it concave at that point. Here is a graphical confirmation:

graph{1 - x - e^x/x^2 [-10, 10, -5, 5]}

Hopefully this helps!

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