Is #f(x)=-x^3-x^2+x# concave or convex at #x=4#?

Question

Isabella Ackerman · Accepted Answer

concave at x = 4

To test if a function is concave/convex at f(a), require to find the value of f''(a) • If f''(a) > 0 then f(x) is convex at x = a • >If f''(a) < 0 then f(x) is concave at x = a hence: f(x) = #-x^3 - x^2 + x # f'(x) = #-3x^2 - 2x + 1# and f''(x) = -6x -2 hence : f''(4) = -6(4) - 2 = - 26 since f''(4) < 0 then f(x) is concave at x = 4

Emma Aldridge · Answer

undefined To determine whether the function $ f(x) = -x^3 - x^2 + x $ is concave or convex at $ x = 4 $, we need to analyze the second derivative of the function at that point.

1. Find the first derivative of $ f(x) $:
$$ f'(x) = -3x^2 - 2x + 1 $$

2. Find the second derivative of $ f(x) $:
$$ f''(x) = -6x - 2 $$

3. Evaluate the second derivative at $ x = 4 $:
$$ f''(4) = -6(4) - 2 = -24 - 2 = -26 $$

Since the second derivative $ f''(4) = -26 $ is negative, the function $ f(x) = -x^3 - x^2 + x $ is concave at $ x = 4 $.

Ezekiel Alexander · Answer

undefined To determine whether $f(x) = -x^3 - x^2 + x$ is concave or convex at $x = 4$, we need to find the second derivative of $f(x)$ and evaluate it at $x = 4$.

The first derivative of $f(x)$ is:
$$f'(x) = -3x^2 - 2x + 1$$

The second derivative of $f(x)$ is:
$$f''(x) = -6x - 2$$

Now, evaluate $f''(x)$ at $x = 4$:
$$f''(4) = -6(4) - 2 = -24 - 2 = -26$$

Since $f''(4) < 0$, the function is concave at $x = 4$.