Expert answers to "Is f(x)=−ex−x3−2x concave or convex at x=−1?"

Angelina Kaufman

Answered question

2022-04-15

f (x) = - \sqrt{e^{x} - x^{3} - 2 x}

concave or convex at

x = - 1

Dallelopeelvep2yc

Beginner2022-04-16Added 15 answers

Step 1

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

Determine the concavity of a graph by finding the second derivative of the function, and the sign of

f (- 1)

epresents the concavity.
Differentiate using the chain rule:

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

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

Step 2
Differentiate using the quotient rule:

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

Plug in

x = - 1

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

This value is negative because the numerator is negative while the denominator is positive.
Therefore, f(x) is comvex at

x = - 1

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