using the graph, determine any relative maxima or minima of the function and the intervals on which the…

Explanation:

Step1: Find the derivative of the function

The derivative of $f(x)=\frac{1}{4}x^{3}-\frac{1}{2}x^{2}-x + 2$ is $f'(x)=\frac{3}{4}x^{2}-x - 1$.

Step2: Set the derivative equal to zero

We solve the equation $\frac{3}{4}x^{2}-x - 1=0$. Multiply through by 4 to get $3x^{2}-4x - 4 = 0$. Factoring gives $(3x + 2)(x - 2)=0$. So $x=-\frac{2}{3}\approx - 0.667$ or $x = 2$.

Step3: Determine the nature of the critical - points

We use the second - derivative test. The second - derivative $f''(x)=\frac{3}{2}x-1$. When $x =-\frac{2}{3}$, $f''(-\frac{2}{3})=\frac{3}{2}\times(-\frac{2}{3})-1=-1 - 1=-2<0$. So the function has a relative maximum at $x =-\frac{2}{3}$. When $x = 2$, $f''(2)=\frac{3}{2}\times2-1=3 - 1 = 2>0$. So the function has a relative minimum at $x = 2$.

Step4: Find the intervals of increase and decrease

We consider the intervals determined by the critical points $x=-\frac{2}{3}$ and $x = 2$. We test values in the intervals $(-\infty,-\frac{2}{3})$, $(-\frac{2}{3},2)$ and $(2,\infty)$. For the interval $(-\infty,-\frac{2}{3})$, let's choose $x=-1$. Then $f'(-1)=\frac{3}{4}+1 - 1=\frac{3}{4}>0$, so the function is increasing on $(-\infty,-\frac{2}{3})$. For the interval $(-\frac{2}{3},2)$, let's choose $x = 0$. Then $f'(0)=-1<0$, so the function is decreasing on $(-\frac{2}{3},2)$. For the interval $(2,\infty)$, let's choose $x = 3$. Then $f'(3)=\frac{27}{4}-3 - 1=\frac{27 - 12 - 4}{4}=\frac{11}{4}>0$, so the function is increasing on $(2,\infty)$.

Answer:

The relative maximum value occurs at $x\approx - 0.667$. The function is increasing on $(-\infty,-\frac{2}{3})\cup(2,\infty)$ and decreasing on $(-\frac{2}{3},2)$. For the multiple - choice part: A. The relative maximum value(s) occur(s) at $x=-0.667$.