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A critical point $c$ is defined as $f'(c) = 0$ or $f'(c) = $ undefined. This definition is taken from this video.

if $$f(x) = x-4\sqrt{x+1}$$ then $$f'(x) = 1 - \frac{2}{\sqrt{x+1}}$$

To find the critical points I set $f'(x) = 0$. According to khanacademy the only critical point is 3. But my answer was $-1$ because if you substitute that for $x$, then $\frac{2}{0}$ would be undefined... Why isn't $-1$ a critical point? I thought undefined was a critical point by definition?

user1534664
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  • I would say you're correct, since $-1$ is in the domain of $f$ and $f'(-1)$ is undefined. Edit: the definition I'm familiar with requires that a critical point be in the domain of the function. $f'$ being undefined is not enough. – Nick D. Apr 04 '15 at 23:07
  • @NickD Quote from wiki: "In mathematics, a critical point or stationary point of a differentiable function of a real or complex variable is any value in its domain where its derivative is 0 or undefined" :) – user1534664 Apr 04 '15 at 23:24

2 Answers2

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Apparently the domain of $f'(x)$ is $[-1, \infty)$, so when $x = -1$ it's at the left endpoint of the domain. Critical points cannot be endpoints.

user1534664
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According to how you define a critical point then both $x=-1$ and $x=3$ are critical points.

I am unfamiliar with the concept of a critical point being one where the derivative is undefined. It seems counter intuitive since a critical point tend to be where the function's rate of change is zero but as $x$ tend to -1 (from above) we have a greater and greater rate of change.

Jai Pancholi
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  • I got this definition from this video https://www.youtube.com/watch?t=23&v=MUQfl385Yug . What's your definition? – user1534664 Apr 04 '15 at 23:13
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    The definition know of is just from the lecturer who has not cited where She pulled it from; she doesn't consider a critical point c where the derivative is undefined. Edit: reading your comment about wiki, I'll question this with her. Thanks – Jai Pancholi Apr 04 '15 at 23:23