Let f be the piecewise defined function given by f(x)=ax+b x≤2. 2-x^2 2<x. Find a and b so that f is differentiable at x=2.
first, remember that if a function is differentiable at $\displaystyle x = c$ , then it is continuous at $\displaystyle x = c$.
secondly ...
$\displaystyle f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$
for your piecewise function ...
$\displaystyle f'(2)^- = f'(2)^+$
$\displaystyle \lim_{x \to 2^-} \frac{(ax+b) - f(2)}{x-2} = \lim_{x \to 2^+} \frac{(2-x^2) - f(2)}{x-2}$
Nevermind. I found that one out too.
But this one is impossible :/
Find the tangent line to the graph of f(x)=1/(x+4) at x=-3.
You will need to use the definition of the derivative to find the slope of the tangent line.
Put your answer in the form y=mx+b.
$\displaystyle f'(x)=-\frac{1}{(x+4)^2}$. Thus $\displaystyle f'(-3)=-1$. Therefore the slope of the perpendicular line will be $\displaystyle 1$, so you want the equation of the line with slope $\displaystyle 1$ going through the point $\displaystyle (-3,1)$.
$\displaystyle y-1=1\cdot(x-(-3)) \implies \boxed{y=x+4}$