Determine where the absolute extrema of f(x)=4x/(x^2+1) on the interval [−4,0] occur

Question

Determine where the absolute extrema of f(x)=4x/(x^2+1) on the interval [−4,0] occur

The absolute maximum occurs at x=

The absolute minimum occurs at x =

Answer 1

First consider how the graph might look. If x = 0, then f(x) also is 0 so it goes through the point (0,0). Also as x> negative infinity, the f(x) goes towards 0. However, values say for x = -1 show that y = -2, so the graph goes through the origin, approaches 0 as x approaches infinity, and displays somewhat of a dip at least early on inside the interval given.

We can conclude graphically that if there is a low, it must be concave up without determining the second derivative.

To find the local max / min, we need to determine where the derivative is 0. Keep in mind though, just demonstrating the derivative is 0 at a point DOES NOT make it a local max or min. One must corroborate though either graphical means, or through taking the second derivative. I have chosen a graphical analysis. 

So if f(x) = 4x/(x^2+1), f’(x) is (4x^2 + 4 – 8x^2)/(x^2+1)^2. We are only concerned about the numerator since we want to know where the entire derivative is 0 – that is where the numerator is zero.

4x^2 + 4 – 8x^2 = 0 if x^2 = 1 or x = 1, or x = -1. -1 is on the interval mentioned, therefore the x value of local minimum must be -1. Solving for y using the original equation determines the point to be (-1,2) as the only local min on the interval.