calculius - functions

Question

The function  has one local minimum and one local maximum.This function has a local maximum at    with value   and a local minimum at    with value   

Answer 1

To find local extrema, take the derivative and set it equal to zero:

For f(x) = 4x + 6x^-1, f '(x) = 4 - 6x^-2

0 = 4 - 6x^-2

6x^-2 = 4

6/x² = 4/1 [cross multiplying gives:

6 = 4x²

x² = 6/4

x = ±√6/2 (critical points)

An additional critical point is where the derivative DNE which is at x=0

Putting these on a number line it divides that number line into 4 pieces:

and calculating slope on each interval we get:

f '(-2) = 4 - 6(-2)^-2 = 2.5 (positive) meaning the function is increasing on this interval

f '(-1) = 4 - 6(-1)^-2 = -2 (negative) meaning the function is decreasing on this interval

f '(1) = 4 - 6(-2)^-2 = -2 (negative) meaning the function is decreasing on this interval

f '(2) = 4 - 6(-2)^-2 = 2.5 (positive) meaning the function is increasing on this interval

Since the function is increasing to the left of x = -√6/2 and decreasing to the right, then x = -√6/2 is a local maximum. The value of that maximum is f(-√6/2) = 4(-√6/2) + 6(-√6/2)^-1 = -4√6

Since the function is decreasing to the left of x = √6/2 and increasing to the right, then x = √6/2 is a local minimum. The value of that minimum is f(√6/2) = 4(√6/2) + 6(√6/2)^-1 = 4√6