Question

the first derivative of the function f is given by f′(x)=cos2xx−15 . How many critical values does f have on the open interval (0,10) ?

Answer 1

Answer:

There are 3 critical values of f on the open interval (0,10).

Step-by-step explanation:

It is given that the first derivative of the function f is  

$f'(x)=\dfrac{\cos^2x}{x}-\dfrac{1}{5}$

We need to find the number of critical values of f on the open interval (0,10).

To find the critical values we have to find the value of x for which f'(x)=0.

$\dfrac{\cos^2x}{x}-\dfrac{1}{5}=0$

$\dfrac{\cos^2x}{x}=\dfrac{1}{5}$

$5\cos^2x=x$

Using graphing calculator we get

$x\approx 1.086, 2.320, 3.681$

All the values 1.086, 2.320, 3.681 lie on the interval (0,10).

Therefore, there are 3 critical values of f on the open interval (0,10).