Question

Is `f(x)=-sqrt(e^x -x^3-2x` concave or convex at `x=-1`?

Answer 1

`f(x)` is convex at `x=-1`.

Explanation:

`f(x)=- sqrt(e^x-x^3-2x) = -(e^x-x^3-2x)^(1/2)`

Determine the concavity of a graph by finding the second derivative of the function, and the sign of `f"(-1)` represents the concavity.

Differentiate using the chain rule:
`f'(x)=-1/2(e^x-x^3-2x)^(-1/2)(e^x-3x^2-2)`

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

Differentiate using the quotient rule:
`f''(x) = frac{-2(e^x-x^3-2x)(e^x-6x)+(e^x-3x^2-2)^2(2)}{4(e^x-x^3-2x)^2}`

Plug in `x=-1`:
`f''(-1)= frac{-2(1/e +1+2)(1/e +6)+(1/e -3 -2)^2(2)}{4(1/e +3)^2}`

This value is negative because the numerator is negative while the denominator is positive.

Therefore, `f(x)` is convex at `x=-1`.