Question

Find absolute maximum and minimum values of a function f given by f(x) = 12x^4/3 – 6x^1/3, x ∈ [–1, 1].

Answer 1

We have,

f(x) = \(12x^\frac{4}{3}\)– \(6x^\frac{1}{3}\)

Thus,

f’(x) = 0

⇒ x = \(\frac{1}{8}\)

Further note that f’(x) is not define at x = 0. 

So, the critical points are x = 0 and x = \(\frac{1}{8}\)and at the end point of the interval x = – 1 and x = 1

Thus, 

We conclude that the absolute maximum value of f is 18 at x = 1, and absolute minimum value of f is \(-\frac{9}{4}\) which occurs at x = \(\frac{1}{8}\).