Question

question. consider the function f(x) below. over what open interval(s) is the function increasing and concave up? give your answer in interval notation. enter ∅ if the interval does not exist. f(x)=x³ - x² - 8x - 10. provide your answer below.

Explanation:

Step1: Find the first - derivative

First, find the derivative of $f(x)=x^{3}-x^{2}-8x - 10$. Using the power rule $(x^n)'=nx^{n - 1}$, we have $f'(x)=3x^{2}-2x - 8=(3x + 4)(x - 2)$.

Step2: Find the critical points

Set $f'(x)=0$. Then $(3x + 4)(x - 2)=0$. Solving for $x$, we get $x=-\frac{4}{3}$ and $x = 2$.

Step3: Find the second - derivative

Differentiate $f'(x)=3x^{2}-2x - 8$ to get $f''(x)=6x-2$.

Step4: Find the inflection point

Set $f''(x)=0$. Then $6x-2 = 0$, so $x=\frac{1}{3}$.

Step5: Test intervals for increasing and concave - up

Test intervals for $f'(x)$ and $f''(x)$.
For $f'(x)$:

• Consider the intervals $(-\infty,-\frac{4}{3})$, $(-\frac{4}{3},2)$ and $(2,\infty)$.
• For $x\in(-\infty,-\frac{4}{3})$, let $x=-2$, then $f'(-2)=3\times(-2)^{2}-2\times(-2)-8 = 12 + 4-8=8>0$.
• For $x\in(-\frac{4}{3},2)$, let $x = 0$, then $f'(0)=-8<0$.
• For $x\in(2,\infty)$, let $x = 3$, then $f'(3)=3\times3^{2}-2\times3 - 8=27-6 - 8 = 13>0$.

For $f''(x)$:

• Consider the intervals $(-\infty,\frac{1}{3})$ and $(\frac{1}{3},\infty)$.
• For $x\in(-\infty,\frac{1}{3})$, let $x = 0$, then $f''(0)=-2<0$.
• For $x\in(\frac{1}{3},\infty)$, let $x = 1$, then $f''(1)=6\times1-2 = 4>0$.

The function is increasing when $f'(x)>0$ and concave - up when $f''(x)>0$.
The interval where the function is increasing and concave - up is $(2,\infty)$.

Answer:

$(2,\infty)$