0

I'm trying to understand an explanation in Khan Academy on the topic "Finding the critical point".

I understand that we need to get the derivative down to either zero or undefined to find that critical point, but I don't get why we can't use use 1 in this equation's derivative. $h(x) = \frac{4x^2}{x^2-1}$ enter image description here

The derivative is $\frac{-8x}{(x^2-1)^2}$

Is it because the graph itself already has undefined at 1? enter image description here

nvs0000
  • 681
  • 1
    A critical point has to be a point on the domain of the function. In other words the function should be defined at that point. Your function is not defined at 1 or -1. – Ameet Sharma Nov 14 '20 at 00:17
  • So is 0 a critical point of a function like this $h(x) = 6x^5-15x^4+10x^3$? I feel like this critical point piece is tripping me up. I've had to ask another question herelink – nvs0000 Nov 14 '20 at 11:48
  • 1
    Yes, in that situation 0 is a critical point, because $h(0)$ is well-defined, and $h'(0)=0$ – Ameet Sharma Nov 14 '20 at 12:48
  • In this videolink, Khan Acad defined critical points as: "non-endpoint min or max, @ x=a, where f'(a) = 0 or f'(a) = undefined." Please help me understand why that "undefined point in Khan Acad link" is a critical point while the "undefined point in $\frac{−8}{(^2−1)^2}$ " is not a critical point. //

    I can see the graphs look different, but.. is that all we are basing it on, we have to draw the graphs to know?

     – nvs0000 Nov 14 '20 at 13:06
  • 1
    if f(x) is well defined but f'(x) is not well defined, then it is a critical point. The key is that f(x) is always well defined at all critical points. But f'(x) need not be. If f(a) is well defined, but f'(a) is undefined, then a is a critical point. In your link f(a) is well defined, because the function is well defined at all x values. So if f'(a) =0 or f'(a) is undefined then a is a critical point, because we know f(a) is well defined. – Ameet Sharma Nov 14 '20 at 13:24
  • 1
    Thanks so much, this comment help! To be a critical point, the selected range in function f must be defined. It doesn't matter if the derivatives are undefined. If the original f(x) is not defined, it cannot be a critical point. – nvs0000 Nov 15 '20 at 04:22

2 Answers2

3

A critical point is a point in the domain of the function where the derivative is either undefined or zero. As you notice, $-1$ and $1$ are not in the domain of the function, so they don't meet the criteria.

We would say that $x=-1$ and $x=1$ are asymptotes of the function.

Rhys Hughes
  • 13,103
  • An asymptote is not the same as undefined? – nvs0000 Nov 14 '20 at 06:56
  • So is 0 a critical point of a function like this $h(x) = 6x^5-15x^4+10x^3$? I feel like this critical point piece is tripping me up. I've had to ask another question herelink – nvs0000 Nov 14 '20 at 11:49
  • 1
    @olif9837 An asymptote means that the function is undefined there. For a vertical asymptote, the function approaches either infinity or negative infinity, which are not numbers in the standard definition of calculus. Since the function does not attain a value at that point, it is undefined. – Toby Mak Nov 14 '20 at 11:57
  • In this videolink, Khan Acad defined critical points as: "non-endpoint min or max, @ x=a, where f'(a) = 0 or f'(a) = undefined." Please help me understand why that "undefined point in Khan Acad link" is a critical point while the "undefined point in −8(2−1)2 " is not a critical point. // I can see the graphs look different, but.. is that all we are basing it on, we have to draw the graphs to know? – nvs0000 Nov 14 '20 at 13:28
  • @olif9837 $a$ is a critical point of an arbitrary function $f(x)$ if and only if $a\in\mathcal D(f)$ (i.e., $f(a)=\text{some number}$) and $f'(a)$ is zero or undefined. With your $h(x)=6x^5-15x^4+10x^3$, $h(0)=0$ (i.e. it is defined), and $h'(0)=0$ too, so it meets both standards. With $f(x)=\frac{4x}{x^2-1}; f(1)$ is not defined as $1\not\in\mathcal D(f)$ and so it doesnt meet the first standard. – Rhys Hughes Nov 20 '20 at 07:43
1

If you have

$h(x) = \frac{4x^2}{x^2-1}$

What is $h(1)$ ? You have something divided by 0 which is undefined.

So 1 cannot be a critical point.

But if you have

$h(x) = |x|$

then $x=0$ is a critical point, because $h(0) = 0$ is well defined and $h'(0)$ is not well defined. So the point is that h is well defined at all critical points. $h'(0)$ can be undefined, but $h(0)$ should be well defined.

Another function:

$f(x) = \begin{cases} 0, & \text{if}\ x=0 \\ \frac{1}{x}, & \text{otherwise} \end{cases}$

This function is well defined everywhere. But $f'(0)$ is not well defined. So 0 is critical point here.

Another example:

$h(x) = \begin{cases} 0, & \text{if}\ x=1 \\ \frac{4x^2}{x^2-1}, & \text{otherwise} \end{cases}$

Here 1 is a critical point because now $h(1)$ is well-defined and $h'(1)$ is not well defined. So by making sure the function is defined at 1, we've made it a critical point now.

Ameet Sharma
  • 2,997