Answer the following questions about the function whose derivative is f'(x)=x^2(x-2)/x+5,x≠-5

Question

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

a) To find critical points, we set the derivative equal to zero and solve for x. In this case, we have

f'(x)=x^2*(x-2)/(x+5), so

x^2*(x-2)/(x+5) = 0 [multiply both sides by x+5 to cancel out the denominator]

x^2*(x-2) = 0 [both factors are set equal to zero, due to the zero product rule]

x^2 = 0 and (x-2) = 0

x = 0 and x = 2 [these are your critical values]

b) To find intervals of increasing and decreasing, I would create a number line with the critical values and asymptotes marked.

<-----(-5)-----0-----2----->

Choose a number in each section, i.e. -6 for less than -5, -1 for between -5 and 0, 1 for between 0 and 2, and 3 for greater than 2.

Plug each of these numbers into the derivative and pay attention to whether they are positive or negative:

f'(-6) = 288 [positive]

f'(-1) = -0.75 [negative]

f'(1) = -0.167 [negative]

f'(3) = 1.125 [positive]

When f'(x) is negative, f(x) is decreasing, and when f'(x) is positive, f(x) is increasing. Our intervals of

decreasing are (-5, 0) and (0, 2), since that is where f'(x) is negative. Then our intervals of increasing are (-∞, -5) and (2, ∞), because that is where f'(x) is positive.

c) To find max/min, you can add the positive/negative signs from b) to your number line:

<--[pos]---(-5)--[neg]---0--[neg]---2--[pos]---> [I draw + and - signs above the number line on paper]

Maximums mean that the derivative is moving from positive to negative, so we want to look for x-values that do that. The only one we have is -5, but since that is an asymptote, it cannot be a maximum or a minimum. Therefore, f(x) does not have a maximum.

Similarly, minimums mean that the derivative is moving from negative to positive, so we want x-values that move like that. Looking at our number line, x = 2 does this, because we know that f(1) is negative and f(3) is positive. Therefore, x = 2 is a local minimum.

I hope this helps!