1
$\begingroup$

enter image description here

In the above question f'(x) is zero at x = 0 , discontinuous at x = 1 and not differentiable at x = 2 . Therefore, the function must have 3 critical points but the answer is A that is only 1 critical point.

Please explain the reason behind why only 1 critical point.

  • 0
    I am afraid that the definition of a critical point is not always the same in different parts of the world. Also, how can $f'$ be zero at $1$? – 2017-02-15
  • 1
    critical point must be in the domain. What is not true for $x=\pm 1$ – 2017-02-15
  • 0
    @imranfat because the critical point is $x=0$. – 2017-02-15
  • 0
    @imranfat sorry, f'(x) = 0 at x = 0. I have edited it – 2017-02-15
  • 0
    How do arrive at the 'Therefore' conclusion? (It's not differentiable at $x=\pm 2$, by the way) – 2017-02-15
  • 0
    @Thomas then how to conclude the number of critical points. – 2017-02-15
  • 0
    If it's the number of zeros of $f^\prime$, then by counting these. You arrive at $1$. If it's the joint number of minima, maxima and zeros of $f^\prime$ you need a different reasoning. But I still don't get how you arrive at $3$. – 2017-02-15
  • 0
    @Thomas My desmos screwed up... – 2017-02-15
  • 0
    @Arnaldo thanks. But then also we have 2 critical points those are 4 and 0 – 2017-02-15
  • 0
    @user411518: That depends on the way how you define critical point. As far as I know, the most common definition is: the roots of $f'(x)=0$. – 2017-02-15
  • 0
    Why should $4$ be a critical point??? You have now claimed several times that there has to be a certain (varying) number of critical points but are not willing or able to provide _any_ reason for this. Are you rollng dice? – 2017-02-15
  • 0
    @Thomas ohk. I got it. – 2017-02-15

1 Answers 1

1

In standard modern Calculus textbooks (at least the ones commonly used in the United States), a critical point is a point of the domain where the derivative is either zero or doesn't exist. Thus there are two types of critical points.

For the function $\displaystyle{f(x) = \frac{\left|x^2-4\right|}{x^2-1}}$

  • $f^\prime(x) = 0\,$ only at $x = 0$.
  • The domain of $f$ is $\{x \in \mathbb{R} \mid x \ne \pm 1\}$.
  • The only values in the domain of $f$ where $f^\prime(x)$ fails to exist are the values $x = \pm 2$.

Therefore $f$ has $3$ critical points, namely $x = 0,\; x = \pm 2$.