"Is f(x)=ex−1−x2x−1 concave or convex at x=−1?" was answered by students like you

Marshall Wolf

Answered question

2022-04-16

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

concave or convex at

x = - 1

Kaphefsceasiaf8w9

Beginner2022-04-17Added 9 answers

Step 1
The derivative of a quotient is

{(\frac{u}{v})}^{'} = \frac{u^{'} v - u v^{'}}{v^{2}}

The first derivative is

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

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

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

Step 2
The second derivative is

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

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

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

x = - 1

f - 1) = e^{- 1 - 1} - \frac{2}{{(- 1 - 1)}^{3}} = 0.135 + 0.25 = 0.385

f - 1) > 0

, we conclude that f(x) is convex at

x = - 1

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