How do you find all local maximum and minimum points given #y=3x^4-4x^3#?

1 Answer
Jan 3, 2017

Answer:

#x=1# is a local minimum.
#x=0# is an inflection point.

Explanation:

First we find the critical values for #f(x)# equating the first derivative to zero:

#f'(x) =12x^3-12x^2 = 0#

#12x^2(x-1) = 0 => x_1=0, x_2=1# are the critical points.

Now we calculate the second derivative:

#f''(x) = 36x^2-24x = 12x(3x-2)#

So we have #f''(1) = 12 >0# and #x_2=1# is a local minimum.
Instaed we have #f''(0) = 0#, so to determine the nature of #x_1=0# we can look at the sign of the first derivative and we can see that:

#f'(x) = 12x^2(x-1) < 0# around #x=0#, so that #x_1#cannot be a local maximum or minimum and is instead an inflection point.

graph{3x^4-4x^3 [-2.813, 2.812, -1.406, 1.407]}