Question

Find the absolute extrema of the following functions on the given closed interval.

f(x) = `6x^(4/3) - 3x^(1/3) ; [-1, 1]`

Answer 1

f(x) = `6x^(4/3) - 3x^(1/3)`

f'(x) = `4/3 6x^(1/3) - 3 1/3 x ^(-2/3)`

f'(x) = 0

⇒ `8x^(1/3) - x^(- 2/3)` = 0

Critical point x = `1/8` ∈ (- 1, 1)`

End points x = – 1, 1

Now f(– 1) = 6 + 3 = 9

`"f"(1/8) = 6(1/16) - 3(1/2)`

= `3/8 - 3/2`

= `(3 - 12)/8`

= `- 9/8`

f(1) = 6 – 3 = 3

Absolute maximum f(– 1) = 9

Absolute minimum `"f"(1/8) = - 9/8`