Can someone help me with this??

Question

Find the critical points (critical x values) for the function f(x) = 3x^2-7x^1/3+4?

Answer 1

The function is f(x) = 3x² - 7x^1/3 + 4.

Differentiate with respect to x.

f'(x) = 6x - (7/3)x^-2/3.

To find the critical numbers f'(x) = 0.

6x - (7/3)x^-2/3 = 0

x[6 - (7/3)x^-1/3] = 0

x = 0 and 6 - (7/3)x^-1/3 = 0

x = 0 and 6x^1/3 = 7/3

x = 0 and x = (7/18)³ = 0.0588.

f(0) = 3(0)² - 7(0)^1/3 + 4 = 4

f((7/18)³) = 3((7/18)³)² - 7((7/18)³)^1/3 + 4 = = 3(7/18)⁶ - 7(7/18) + 4 = 1.288.

The critical points are (0, 4) and (0.06, 1.29).