Question

Answer the following questions about the function whose derivative is f primef′​(x)equals=x Superscript negative three fifths Baseline (x minus 3 )x− 3 5(x−3). a. What are the critical points of​ f? b. On what open intervals is f increasing or​ decreasing? c. At what​ points, if​ any, does f assume local maximum and minimum​ values?

Answer 1

Answer:

(a) The critical points of f are x=0 and x=3.

(b)f is decreasing on $(0,3)$ and f is decreasing on $(3,\infty)$.

(c) Therefore the local minimum of f is at x=3

Step-by-step explanation:

Given function is

$f'(x)= x^{-\frac35}(x-3)$

(a)

To find the critical point set f'(x)=0

$\therefore x^{-\frac35}(x-3)=0$

$\Rightarrow x=0,3$

The critical points of f are 0,3.

(b)

The interval are $(0,3)$ and $(3,\infty)$.

To find the increasing or decreasing, taking two points one point from the interval (0,3) and another point $(3,\infty)$.

Assume 1 and 4.

Now $f'(1)=(1)^{-\frac35}(1-3)<0$

and $f'(4)=(4)^{-\frac35}(4-3)>0$

Since 1∈$(0,3)$ , f'(x)<0  and 4∈$(3,\infty)$ , f'(x)>0

∴f is decreasing on $(0,3)$ and f is decreasing on $(3,\infty)$.

(c)

$f'(x)= x^{-\frac35}(x-3)$

Differentiating with respect to x

$f''(x)=-\frac35x^{-\frac 85}(x-3)+x^{-\frac35}$

Now

$f''(0)=-\frac35(0)^{-\frac 85}(0-3)+(0)^{-\frac35}=0$

and

$f''(3)=-\frac35(3)^{-\frac 85}(3-3)+3^{-\frac35}$

        $=0.517>0$

Since f''(x)>0 at x=3

Therefore the local minimum of f is at x=3