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

Question

1761 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-17T12:29:45

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

No calculus answer

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

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

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

Using the derivative

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

$y' = 12x^2$ 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.)

How do you find the local extrema for y=4x^3 + 7y=4x3+7y=4x^3 + 7?