Respuesta :
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
[tex]f'(x)=\dfrac{\cos^2x}{x}-\dfrac{1}{5}[/tex]
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.
[tex]\dfrac{\cos^2x}{x}-\dfrac{1}{5}=0[/tex]
[tex]\dfrac{\cos^2x}{x}=\dfrac{1}{5}[/tex]
[tex]5\cos^2x=x[/tex]
Using graphing calculator we get
[tex]x\approx 1.086, 2.320, 3.681[/tex]
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).
The total number of critical values does f have on the open interval (0,10) is 3 and this can be determined by using the given data.
Given :
[tex]\rm f'(x)=\dfrac{cos^2x}{x}-\dfrac{1}{5}[/tex]
The following steps can be used in order to determine the total number of critical values does f have on the open interval (0,10):
Step 1 - Write the given differential function.
[tex]\rm f'(x)=\dfrac{cos^2x}{x}-\dfrac{1}{5}[/tex]
Step 2 - The critical value is obtained when the differential function is zero.
[tex]\rm0=\dfrac{cos^2x}{x}-\dfrac{1}{5}[/tex]
Step 3 - Simplify the above expression.
[tex]\rm \dfrac{cos^2x}{x}=\dfrac{1}{5}[/tex]
[tex]\rm 5cos^2x = x[/tex]
Step 4 - The value of 'x' using the graphical calculator are:
x [tex]\approx[/tex] 1.086, 2.320, 3.681
So, the total number of critical values does f have on the open interval (0,10) is 3.
For more information, refer to the link given below:
https://brainly.com/question/24062595
