Question

Find the maximum and the minimum point of Y= -(X+1)^3 +8

Answer 1

The derivative of $y$ is $y'(x) = -3 (x+1)^2$. Find critical points, by equating it to zero: $-3 (x+1)^2 = 0$. The last equation has root $x = -1$. The sign of the the derivative in both intervals $(-\infty, -1)$ and $(-1, \infty)$ is negative, so $x= -1$ is not a relative extremum point. The range of function is $\mathbb{R}$, so it doesn't have a maximum or minimum.