1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $x>0$.

$\endgroup$
1
$\begingroup$

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

Nevertheless, they are turning points.

enter image description here

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.