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Is there ever a time when there are no critical points of a function?

For example, I am trying to find the critical points and the extrema of $\displaystyle f(x)= \frac{x}{x-3}$ in $[4,7]$

I am not finding any critical points in this, is this correct?

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the Weierstrass function – Gahawar Jun 28 '14 at 3:28
@Gahawar: Wouldn't that function have a critical point everywhere? Since there exists no derivative? – Dair Jun 28 '14 at 3:31
In the calculus textbook I used, critical points were any point $x = c$ such that $f'(c) = 0$ or $f'(c)$ does not exist. By this definition, the Weierstrass function has critical points at every real number. – JimmyK4542 Jun 28 '14 at 3:31
@JimmyK4542 The definition that I am familiar with is the former condition. – Gahawar Jun 28 '14 at 3:33
The terminology isn't standardized. Some authors only include domain points where the derivative vanishes, some also include domain points where the derivative doesn't exist, and some include boundary points. (This is all in the context of real-valued functions of a real variable.) I learned it as the "intermediate" definition: a critical point is a domain point where the derivative vanishes or fails to exist. Then local extrema occur at critical points or boundary points. – MPW Jun 28 '14 at 3:41

Plenty. For example $f(x)=x$ has no critical points. Neither does $f(x)=e^x$.

And your function has no critical points, according to many definitions. Some definitions would include endpoints among the critical points. In that case, if we consider the function as having domain $[4,7]$, you have $4$ and $7$.

The function is continuous on the interval $[4,7]$. So it does attain a maximum and a minimum in the interval. Since the derivative is defined everywhere in the interval, and nowhere $0$, the endpoints $4$ and $7$ are your only candidates for maximumhood/minimumhood.

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Note that $f(x) = \dfrac{x}{x-3} = 1 + \dfrac{3}{x-3}$. So, $f'(x) = -\dfrac{3}{(x-3)^2} < 0$ for all $x \neq 3$,

Hence, there are no critical points on $(4,7)$.

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