Expert answers to "Is f(x)=1−x−exx2 concave or convex at x=−1?"

Dakota Livingston

Answered question

2022-04-14

f (x) = 1 - x - \frac{e^{x}}{x^{2}}

concave or convex at

x = - 1

tutaonana223a

Beginner2022-04-15Added 15 answers

Step 1
The second derivative is determinative of concavity. Start by finding the first derivative.

f^{'} (x) = - 1 - \frac{e^{x} (x^{2}) - e^{x} (2 x)}{{(x^{2})}^{2}}

f^{'} (x) = 1 - \frac{x (x e^{x} - 2)}{x^{4}}

f^{'} (x) = 1 - \frac{x e^{x} - 2}{x^{3}}

Now differentiate this.

f x) = 0 - \frac{(e^{x} + x e^{x}) x^{3} - (x e^{x} - 2) (3 x^{2})}{{(x^{3})}^{2}}

f x) = - \frac{x^{3} e^{x} + x^{4} e^{x} - 3 x^{3} e^{x} + 6 x^{2}}{x^{6}}

f x) = - \frac{x^{4} e^{x} - 2 x^{3} e^{x} + 6 x^{2}}{x^{6}}

Step 2
We now evaluate the point within the second derivative.

f - 1) = - \frac{{(- 1)}^{4} e^{- 1} - 2 {(- 1)}^{3} e^{- 1} + 6 {(- 1)}^{2}}{{(- 1)}^{6}}

f - 1) = - \frac{\frac{1}{e} - 2 \frac{- 1}{e} + 6}{1}

f - 1) = - \frac{\frac{1}{e} + \frac{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.

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