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I have the equaition $f(x)=x^2(1-x)^4$.

It's derivative is $f'(x)=(1-x)^3(2x-6x^2)$ (as I calculated), and it's second derivative is $f''(x)=(1-x)^2(24x^2-14x+2)$ (again, as I calculated).

When finding where $f'(x)=0$, you get 3 $x$ values: $x_1=0, x_2=1, x_3=\frac{1}{3}$.

When putting them into the second derivative, you get: $$ f''(0)=2 \\ f''(1) = 0 \\ f''(\frac{1}{3})=0 $$ From this, I thought the only extreme point of $f(x)$ was at $x=0$. However, the points at $x = 1 \ and \ x=\frac{1}{3}$ are also extreme points.

Why is that? How can I find thoes points? Thanks for your help.

Guy Adler
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1 Answers1

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Every point $t$ where $f'(t)=0$ is an extreme or turning point.

The value of $f''(t)$ simply determines which type:

  1. $f''(t)>0$ gives a trough, or local minimum
  2. $f''(t)=0$ gives a point of inflection
  3. $f''(t)<0$ gives a peak or local maximum
Rhys Hughes
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  • I understanf your answer, but from my answers to the question, (1, 0) is a local minimum and (1/3, 16/729) is a local maximum. Do you know a reason this might be, or is it just a mistske? – Guy Adler Sep 21 '19 at 16:17
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  • can also be a flat point, this is the case here for $x=1$ making it a minimum. When derivatives are zero, you need to further derivatives to have a definitive answer.
    • – zwim Sep 21 '19 at 17:04