Question

Consider the following function.? f(x) = (x + 3)^2/3

Answer 1

Critical number are obtained by finding the first derivative.

By the chain rule;

`f'(x) = 1 * 2/3(x + 3)^(2/3 - 1)`

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

Critical points will occur when the derivative equals `0`.

`0= 2/3(x+ 3)^(-1/3)`

Clearly `x = -3` is the only value of `x` that satisfies.

If we test `x = 0` in the derivative, we see that we will get a positive result. However, if we use `x= -6` we get a negative number. Therefore, the function is decreasing on `(-oo, -3)` and increasing on `(-3, oo)`. This also means that `x= -3` is a minimum. Furthermore, since the function is continuous on all `x`, and there is only one critical point, there will be no maximum.

Thus, the relative minimum will occur at `f(-3) = (-3 +3)^(2/3) = 0`, so the point is `(-3, 0)`.

We finish with a graphical confirmation.

Hopefully this helps!