I have a piecewise function
$$f(x)=\begin{cases}-x^3-4x^2+10, & x < -2\\ ax+b ,& x \geq -2\end{cases}$$
I need to find all the values of a and b that make $f(x)$ both continuous and differentiable at $x=-2$
I am confused how to go about solving this. what I did was plug the two equations equal to each other and got
$$2=-2a+b$$
Now I don't know what to do with this information. How can I proceed and finish the question?
We have a piecewise function: $$ f(x) = \begin{cases} -x^3-4x^2+10 \quad &x<-2 \\ ax+b &x\geq -2 \end{cases} $$ Notice that for $x<-2$ and for $x\gt -2$ the function is continous and differentiable, since it is the sum of continous and differentiable functions.
What about for $x=-2$? We want both the continuity and the differentiability of $f(x)$ at that point. Therefore, the conditions to be satisfied are:
For continuity: $$ f(-2)=\lim_{x\to -2^-}f(x)=\lim_{x\to -2^+}f(x) $$ that is $$-2a+b=2$$
For differentiability: $$ \lim_{h\to 0^-}\frac{f(-2+h)-f(-2)}{h}=\lim_{h\to 0^+}\frac{f(-2+h)-f(-2)}{h} $$ that is $$ -3x^2-8x=a \\ a=4 $$
Therefore, we get: $$ a=4, b=10 $$