13 mark for review let f be the function defined by f(x) = 1/4 x^4 - 2/3 x^3 + 1/2 x^2 - 1/2 x. for how many…

Explanation:

Step1: Find the average rate of change

The average rate of change of $y = f(x)$ on $[a,b]$ is $\frac{f(b)-f(a)}{b - a}$. Here, $a = 0$, $b=1.565$, and $f(x)=\frac{1}{4}x^{4}-\frac{2}{3}x^{3}+\frac{1}{2}x^{2}-\frac{1}{2}x$. First, find $f(0)$ and $f(1.565)$. $f(0)=\frac{1}{4}(0)^{4}-\frac{2}{3}(0)^{3}+\frac{1}{2}(0)^{2}-\frac{1}{2}(0)=0$. $f(1.565)=\frac{1}{4}(1.565)^{4}-\frac{2}{3}(1.565)^{3}+\frac{1}{2}(1.565)^{2}-\frac{1}{2}(1.565)$. The average - rate of change is $\frac{f(1.565)-f(0)}{1.565 - 0}=\frac{f(1.565)}{1.565}$.

Step2: Find the instantaneous rate of change

The instantaneous rate of change of $y = f(x)$ is $f'(x)$. Differentiate $f(x)$ using the power rule $(x^n)'=nx^{n - 1}$. $f'(x)=x^{3}-2x^{2}+x-\frac{1}{2}$.

Step3: Set them equal and solve

Set $f'(x)$ equal to the average rate of change $\frac{f(1.565)}{1.565}$. Let $g(x)=x^{3}-2x^{2}+x-\frac{1}{2}-\frac{f(1.565)}{1.565}$. We can use the Intermediate - Value Theorem and analyze the behavior of $g(x)$ on the interval $(0,1.565)$. Another way is to use the Mean - Value Theorem. The Mean - Value Theorem states that if $y = f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists at least one $c\in(a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b - a}$. We can also graph $y = f'(x)$ and the horizontal line $y=\frac{f(1.565)-f(0)}{1.565 - 0}$ on a graphing utility. By graphing $y = f'(x)=x^{3}-2x^{2}+x-\frac{1}{2}$ and the horizontal line representing the average rate of change (after calculating $\frac{f(1.565)}{1.565}$) or by analyzing the function $g(x)$ and its derivatives, we find that the number of solutions of $f'(x)=\frac{f(1.565)-f(0)}{1.565 - 0}$ in the open interval $(0,1.565)$ is three.

Answer:

C. Three