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

Question

1533 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-17T12:29:45

#y=4x^3+7# has no local extrema.

No calculus answer

#y = 4x^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 = 4x^3+7#.

Using the derivative

The domain of #4x^3+7# is #(-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.)