Let f(x) = 3-x^2, x _<2 (here x is less than or equal to 2)
4x-9, x<2
Determine whether the graph of y=f(x) has a tangent at x=2. If not, explain why not.
$\displaystyle f(x)=\left\{\begin{array}{ll}3-x^2, & x\geq 2\\4x-9, & x>2\end{array}\right.$
$\displaystyle \lim_{x\nearrow 2}f(x)=\lim_{x\searrow 2}f(x)=f(2)=-1\Rightarrow$ f continuous in x=2.
f is differentiable in a vicinity of x=2.
$\displaystyle f'(x)=\left\{\begin{array}{ll}-2x, & x<2\\4, & x>2\end{array}\right.$
$\displaystyle f_{left}'(2)=\lim_{x\nearrow 2}f'(x)=-4$
$\displaystyle f_{right}'(2)=\lim_{x\searrow 2}f'(x)=4$
So f is not differentiable in x=2 and the graph has not a single tangent. The graph has two semitangents (x=2 is an angular point)