Question
Consider the function $f(x)$ below. Over what open interval(s) is the function decreasing and concave down? Give your answer in interval notation.
\[
f(x)=\frac{x^{3}}{3}+\frac{5 x^{2}}{2}-50 x-6
\]

Enter $\varnothing$ if the interval does not exist.

Provide your answer below:
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Step 1 ：Find the derivative of the function $f(x)$: $f'(x) = x^{2} + 5x - 50$

Step 2 ：Find the second derivative of the function $f(x)$: $f''(x) = 2x + 5$

Step 3 ：Set $f'(x) = 0$ to find the values of $x$ where the function may change from increasing to decreasing or vice versa: $x^{2} + 5x - 50 = 0$

Step 4 ：Factor the equation to find the roots: $(x - 5)(x + 10) = 0$, so $x = 5$ or $x = -10$

Step 5 ：Set $f''(x) = 0$ to find the values of $x$ where the function may change from concave up to concave down or vice versa: $2x + 5 = 0$, so $x = -\frac{5}{2}$

Step 6 ：Test the intervals $(-\infty, -10)$, $(-10, -\frac{5}{2})$, $(-\frac{5}{2}, 5)$, and $(5, \infty)$ to see where $f'(x) < 0$ and $f''(x) < 0$

Step 7 ：For $x < -10$, $f'(x) > 0$ and $f''(x) > 0$

Step 8 ：For $-10 < x < -\frac{5}{2}$, $f'(x) > 0$ and $f''(x) < 0$

Step 9 ：For $-\frac{5}{2} < x < 5$, $f'(x) < 0$ and $f''(x) < 0$

Step 10 ：For $x > 5$, $f'(x) > 0$ and $f''(x) > 0$

Step 11 ：Therefore, the function is decreasing and concave down on the interval $(-\frac{5}{2}, 5)$

Step 12 ：\(\boxed{(-\frac{5}{2}, 5)}\) is the interval where the function is decreasing and concave down