Derivatives and Shape of Graph

Question

Let p(z) = 5z^4-10z^2+15. Find the values or intervals below.

interval(s) on which p is increasing:

interval(s) on which p is decreasing:

point(s) at which p achieves a local maximum:

point(s) at which p achieves a local minimum:

interval(s) on which p concaves up:

interval(s) on which p concaves down:

Answer 1

p(z) is increasing when p'(z) is positive and decreasing when p'(z) is negative. p'(z) = 20z³ - 20z = 20z(z² -1) = 20z(z + 1)(z - 1). Setting this equal to zero, we find critical points which could be places that are boundaries where the changes can occur.

20z(z + 1)(z - 1) = 0 when z = 0, when z = 1, and when z = -1. There are no places where the derivative DNE. Break a number line up using these critical points and try values on each interval:

This first derivative test shows the intervals where p(z) is increasing and decreasing. We can also find the minimums and maximums. Since p(z) in decreasing left of -1 and increasing right of -1 then z = -1 must be a local minimum. Since p(z) is increasing left of zero and decreasing right of zero, z = 0 must be a local maximum. Since p(z) is decreasing left of z = 1 and increasing to the right, z = 1 must be a local minimum.

To find concavity, take the second derivative. p''(z) = 60z² - 20 = 20(3z² - 1) and setting this equal to zero finds possible points of inflection where concavity changes.

20(3z² - 1) = 0

3z² - 1 = 0

3z² = 1

z² = 1/3

z = ±√(1/3)

This divides the number line up into 3 parts. Try values in each section to find the value of p''(z). If p''(z) is positive, then p(z) is concave up. If it's negative, p(z) is concave down. There are no places where p''(z) DNE.

This shows the intervals of concavity.