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

1 Answer
Andrea S.
Jan 3, 2017

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_1cannot 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]}

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