Find the critical numbers and the intervals on which the function 𝑓(𝑥)=5𝑥^5−4𝑥^3+9

Question

Find the critical numbers and the intervals on which the function 𝑓(𝑥)=5𝑥^5−4𝑥^3+9is increasing or decreasing. Use the First Derivative Test to determine whether the critical number is a local minimum or maximum (or neither).

(Use symbolic notation and fractions where needed. Give your answer in the form of comma separated list. Enter NONE if there are no critical points.)

The critical numbers with local minimum: ????

The critical numbers with local maximum: ????

(Use symbolic notation and fractions where needed. Give your answers as intervals in the form (*,*). Use inf for infinity, U for combining intervals, and appropriate type of parenthesis "(", ")", "[" or "]" depending on whether the interval is open or closed.)

The function is increasing on: ????

The function is decreasing on: ????

Answer 1

Critical numbers are where the first derivative = 0 or is undefined.

f'(x) = 25x^4 - 12x^2

f'(x)=0: 25x^4 -12x^2 = 0

u = x^2

25u^2 -12u = 0

u(25u -12)=0

u=0, u = 12/25

x^2 = 0

x = 0

x^2 = 12/25

x = +/- sqrt(12/25) = +/- 2sqrt(3)/5

When f'(x) > 0 it's increasing. When f'(x)<0 it's decreasing.

Intervals: - infinity to -2sqrt(3)/5; -2sqrt(3)/5 to 0, 0 to 2sqrt(3)/5, 2sqrt(3)/5 to infinity.

2sqrt(3)/5 = about .69

- infinity to -2sqrt(3)/5: Use -10: 25(10)^4 -12(10)^2 = positive. Increasing on this interval.

-2sqrt(3)/5 to 0: Use -.5: 25(-.5)^4 - 12(-.5)^2 = negative. Decreasing on this interval.

0 to 2sqrt(3)/5: Will be the same as -.5 because the powers are even = negative. Decreasing on this interval.

2sqrt(3)/5 to infinity: Use 10. Same as -10 because the powers are even = positive. Increasing on this interval.

Local Max is f'(x) is positive before the critical point and negative after the critical point. -2sqrt(3)/5 is x coordinate of a local max.

Local Min is f'(x) is negative before the critical point and positive after the critical point. 2sqrt(3)/5 is x coordinate of a local min.