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

1 Answer
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!