How do I find the point in which the first derivative does not exist (in order to find the extremas) without the use of limits?
Is it correct to find the first derivative and find it's domain and say that the first derivative does not exists in point outside of it's domain?
Thanks
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$\begingroup$There are a few cases in which the first derivative does not exist by inspection of $f(x)$.
Case 1: Discontinuity
Consider the piece-wise function $$f(x) = \left\{\begin{matrix}x & x>1\\ x & x<1\end{matrix}\right. $$ Which has a discontinuity at $x=1$, and therefore the first derivative does not exist (there is no tangent line through a point which does not exist)
Case 2: Asymptote
Consider $$f(x) = \frac{1}{x}$$ At $x=0$, the limit approaches both positive and negative infinity, so the limit does not exist, but neither does the first derivative (by definition). But even in the case where we can evaluate the limit like in the case of $f(x) = \frac{1}{x^{2}}$ where the limit as $x \rightarrow 0$ $f(x)$ goes to $\infty$, the first derivative also does not exist here.
Case 3: Corner/Cusp
Dr. Sonnhard already gave the example for $f(x) = \left | x \right |$ as an example of a corner where the first derivative does not exist. An example of a cusp is $f(x) = x^{\frac{2}{3}}$ which has a cusp at $x=0$, and so the first derivative does not exist at $x=0$
Case 4: Vertical tangent
A vertical tangent is a point on $f(x)$ where the tangent to the point has infinite slope (it is a vertical line). An example of this is $f(x) = \sqrt[5]{2-x}$ which has a vertical tangent at $x = 0$, so the first derivative does not exist.
I must add that if you want to rigorously prove that a derivative does not exist for a given point on the domain, then you should use limits. But there are qualities of a function that can be inspected to know if the limit exists or not. you should graph these functions so you can better visualize why there is no valid first derivative for certain points on these functions for each case.
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