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The first derivative of the function f is given by f'(x)=[(cos^2x)/x]-1/5. How many critical values does f on the open interval (0,10)?

User Riron
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Answer: A critical value of a function f(x) is a point x in the domain of f(x) where either the derivative is equal to zero or the derivative is undefined.

In this case, the derivative of f(x) is given by:

f'(x) = (cos^2(x))/x - 1/5

To find the critical points of f(x) on the interval (0, 10), we need to solve for x when f'(x) = 0 or f'(x) is undefined.

Setting f'(x) equal to zero, we get:

(cos^2(x))/x - 1/5 = 0

(cos^2(x))/x = 1/5

cos^2(x) = x/5

Taking the square root of both sides, we get:

cos(x) = sqrt(x/5)

This equation has solutions on the interval (0, 10) where x/5 is less than or equal to 1, since the range of the cosine function is between -1 and 1. Therefore, we can write:

0 < x/5 <= 1

0 < x <= 5

So we need to find the values of x between 0 and 5 that satisfy the equation cos(x) = sqrt(x/5).

To do this, we can graph the two functions y = cos(x) and y = sqrt(x/5) on the same set of axes and look for their intersection points between 0 and 5.

Using a graphing calculator or a software, we can see that there is only one intersection point between the two functions on the interval (0, 5). This intersection point is approximately x = 0.433.

Therefore, the function f(x) has only one critical point on the interval (0, 10), which is located at x = 0.433.

User Redlightbulb
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