# Given function f(x)=x*(x^2-1)^1/3. If M = maxim{f(x) / -2<=x<=1}, then: a. M=0 b. M=(27/125)^1/6 c.M= (540^1/6)/5 d. M=1

Posted on

To find the maximum value of the function on the specified interval, we need to take its derivative, set that to zero and solve for the critical points.  Since the interval is closed, we can evaluate the function at the endpoints and the critical points to determine the maximum value.

Take the derivative of the function `f(x)=x(x^2-1)^{1/3}`  using the product rule and chain rule.

`f'(x)=(x^2-1)^{1/3}+x(1/3)(2x)(x^2-1)^{-2/3}`  factor

`=1/{3(x^2-1)^{2/3}}(3(x^2-1)+2x^2)`   simplify numerator

`={5x^2-3}/{3(x^2-1)^{2/3}}`

The critical points are found when the derivative is zero (zeros of the numerator) or doesn't exist (zeros of the denominator).

The zeros of the numerator are at `x=pm sqrt{3/5}` which are both in the specified domain.

The zeros of the denominator are at `x=pm 1` which are also in the specified domain.

This gives 5 points to evaluate (including the one end point that is not a critical point).

`f(-2)=-2(3)^{1/3} approx -2.88`

`f(-1)=0`

`f(1)=0`

`f(sqrt{3/5})=sqrt{3/5}(3/5-1)^{1/3}=sqrt{3/5}(-2/5)^{1/3}approx -0.57`

`f(-sqrt{3/5})=-sqrt{3/5}(3/5-1)^{1/3}=sqrt{3/5}(2/5)^{1/3} approx 0.57`  which is the maximum

Note that the maximum value can also be written as

`{3^{1/2}2^{1/3}}/{5^{1/2}5^{1/3}}`

`={3^{3/6}2^{2/6}}/5^{5/6}`

`=(3^3 2^2 5^1)^{1/6}/5`

`=540^{1/6}/5`