How do you find points of inflection for (x) = 2x(x-4)^3?

1 Answer
Shura
May 14, 2015

There are two points of inflection : one when x_1=2 and another when x_2=4.

The points of inflection of a function are given by the study of the sign the second derivative. Those are the points where the concavity of a function changes.

Let's derive f(x) = 2x(x-4)^3 :

f'(x) = (2x)' * (x-4)^3 + ((x-4)^(3))' * (2x)

f'(x) = 2(x-4)^3 + 6x(x-4)^2 = (x-4)^2(2x-8+6x)

f'(x) = 8(x-4)^2(x-1)

Now, let's derive f'(x) :

f''(x) = ((x-4)^2)' * 8 * (x-1) + (x-1)' * 8 * (x-4)^2

f''(x) = 16(x-4)(x-1) + 8(x-4)^2

f''(x) = (x-4)(16x-16+8x-32)

f''(x) = (x-4)(24x-48) = 24(x-4)(x-2)

f''(x) = 0, when x_1=2 and x_2=4.

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Therefore, your function has two points of inflection : one when x_1=2 and another when x_2=4.

That's it!

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