Consider the following function. 𝑓(𝑥) = 𝑥 + 3 / 𝑥^2 .

Question

a) Find the critical numbers of f, if any.

b) Find the open intervals on which the function is increasing or decreasing. (Enter your answers using interval notation. If no answer exists write DNE)

c) Apply the First Derivative Test to identify all relative extrema. . (Enter your answers using interval notation. If no answer exists write DNE)

Answer 1

a) I believe the critical numbers are all the points at which the function's derivative df/dx equals 0 OR is not differentiable. So:

df/dx = 1 - 6/x³ = 0

Where is this 0 or undefined?

b) Once you have found the critical points, check a value of df/dx in between each one and at the outer boundaries of the domain (aka between the largest critical point and infinity and between the lowest critical point and negative infinity). The derivative sign will tell you whether it's an increasing or decreasing function in each part of the domain. And since the function is assumed continuous between the critical points, we expect the sign of the derivative to remain constant until the next critical point.

c) The relative extrema occur at all the critical points where the derivative is ZERO, not where it's discontinuous. You can identify whether it's a maximum or minimum by the order of increasing vs decreasing on each side of the critical point. I also urge you to graph the function to understand your answer; a tool like desmos.com/calculator or a graphing calculator is always handy in these problems!

Answer 2

a) f'(x) = 1 - 6/x³ = 0; x = 6^1/3.

b) f'(x) = 1 - 6/x³ > 0; x < 0 or x > 6^1/3; (- ∞, 0) and (6^1,/3, ∞) are intervals of increasing; (0, 6^1/3) f'(x) < 0 interval of decreasing.

c) At x = 6^1/3 f'(x0 change sign from - to + from left to right, so at x = 6^1/3 we have local minimum f(6^1/3) =

6^1/3+ 3/6^2/3 = 6^1/3 + 6^1/3/2 = 3·6^1/3/2