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Let us consider $f(x) = x |x| - x.$ Find the critical and inflection points.

I have found the critical points by setting $f'(x) = 0.$ It gives $-\frac12, 0, \frac12$ are the critical points. But to find inflection points I set $f''(x) = 0$ and $x = 0$ becomes an inflection point. But graph of this function says that $-\frac12$ and $\frac12$ are also inflection points as $f(x)$ changes sign there. Kindly tell me about this.

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2 Answers 2

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The derivative is $$ f'(x)=2|x|-1 $$ for $x\ne0$. On the other hand, $f$ is continuous at $0$ and $\lim_{x\to0}f'(x)=-1$, so $f$ is also differentiable at $0$ and the above formula also holds for $x=0$.

Thus the critical points are $-1/2$ and $1/2$, respectively a point of maximum and minimum.

The second derivative only exists for $x\ne0$: $$ f''(x)=\frac{|x|}{x} $$ However, $0$ is a point of inflection, because $f''(x)<0$ for $x<0$ and $f''(x)>0$ for $x>0$.

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The curvature doesn't change at $\pm \frac12$, hence they are not inflection points.

Nevertheless, they are turning points.

enter image description here

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  • $\begingroup$ Ohhhoo...sorry sir, I missed the curvature part. So (-1/2, 0, 1/2) are critical points and 0 is inflection point. $\endgroup$
    – Manglu
    Jan 27, 2018 at 17:18
  • $\begingroup$ yup, curvature changes at $0$. $\endgroup$ Jan 27, 2018 at 17:20

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