Answer to Question #332448 in Calculus for godwin

Question

prove that f(x)= | x+1| has no tangent line at (-1 0)

Answer 1

$f(x)=|x+1|,\ \ x_0=-1,\ f(x_0)=0$

$\lim\limits_{h\rightarrow 0+}\frac{f(x_0+h)-f(x_0)}{h}= \lim\limits_{h\rightarrow 0+}\frac{|-1+h+1|-0}{h}= \lim\limits_{h\rightarrow 0+}\frac{h}{h}=1$

$\lim\limits_{h\rightarrow 0-}\frac{f(x_0+h)-f(x_0)}{h}= \lim\limits_{h\rightarrow 0-}\frac{|-1+h+1|-0}{h}= \lim\limits_{h\rightarrow 0-}\frac{-h}{h}=-1$

So, $\lim\limits_{h\rightarrow 0}\frac{f(x_0+h)-f(x_0)}{h}$ doesn’t exist and isn’t either $\infty$ or $-\infty$ .

Then the graph of $f$ has no tangent line at $(-1,\ 0)$ .