# How do you find the local extrema for y=4x^3 + 7?

May 17, 2016

$y = 4 {x}^{3} + 7$ has no local extrema.

#### Explanation:

$y = 4 {x}^{3} + 7$ is an elongated and translated graph based on $y = {x}^{3}$.

(Multiplying by $4$ stretches the graph vertically and adding $7$ translated the graph upward by $7$.)

Since $y = {x}^{3}$ has no extrema, neither does $y = 4 {x}^{3} + 7$.

Using the derivative

The domain of $4 {x}^{3} + 7$ is $\left(- \infty , \infty\right)$

$y ' = 12 {x}^{2}$ is never undefined and is $0$ only at $0$, so

the only critical number is $0$.

But $y '$ is positive for all values of $x$, so (by the first derivative test) there is neither a minimum nor a maximum at $x = 0$.

(There are no other critical numbers to consider, so there is no extremum.)