Question

Pleas can you help to? find the absolute minimum and maximum of f(x)=x^2/3 0n [-2,3]?

Answer 1

Absolute minimum: `f(x=0)=0`
Absolute maximum: `f(x=3)=root(3)(9)`

Explanation:

The absolute minimum and maximum values must occur at critical points for the function (if such points are within the given range) or at the end points of the given range.

Critical points occur where the derivative of the function is equal to zero.
`f(x)=x^(2/3)`
`color(white)("XXX")rarr f'(x)=2/3x^(-1/3`
with a critical point at
`color(white)("XXX")2/3x^(-1/3)=0`
`color(white)("XXXXXX")rarr x=0`
`f(x=0)=0`

The end points of the given range are
`f(x=-2)=(-2)^(2/3)=root(3)((-2)^2)=root(3)(4)`
and
`f(x=+3)=(+3)^(2/3)=root(3)((+3)^2)=root(3)(9)`

Since `0 < root(3)(4) < root(3)(9)`

the minimum is `0`
and
the maximum is `root(3)(9)`