I have the following problem that I'm not sure how to tackle. For function $f$, find values for $a$ and $b$ such that $f'(3)$ exists.
The function is defined as:
$$ f(x)= \begin{cases} \hfill x^2-4x+8 & \text{if $x \leq 3$} \\ ax+b & \text{if $x \gt 3$} \\ \end{cases} $$
So I'm a bit at a loss here. Both cases of the function are polynomials, so I would assume that they are always derivable, and therefore $f'(3)$ would exist, but I'm not sure how to tackle it. Should I derive both equations and then use the results to create a lineal system of equations? How would you recommend me to proceed?
Also, I'm thinking that maybe $f(x)$ is not continuous in $f(3)$ because the limit approaching by the right might be different from the limit approaching from the left. Does it mean it is not derivable and thus $f'(3)$ actually does not exist?
Continuous piecewise functions composed of differentiable functions are not always differentiable. For example, consider the absolute value function:
$$ |x| = \begin{cases} x & \text{if $x \ge 0$} \\ -x & \text{if $x < 0$} \end{cases} $$
Despite being composed of differentiable functions, the function $f(x) = |x|$ is not differentiable at $x = 0$. Taking the limit from the positive side would indicate the slope is $1$, but taking the limit from the negative side would indicate the slope is $-1$.
For most values of $a$ and $b$, your function will look like this (neither continuous nor differentiable).
When $x^2-4x+8 = ax+b$ at the point $x = 3$, this function will become continuous, but still not necessarily differentiable:
For the function to become differentiable, the functions $f(x)=x^2-4x+8$ and $g(x)=ax+b$ and their derivatives $f'(x)=2x-4$ and $g'(x)=a$ must be equal at the point $x = 3$.
You should be able to solve it from there.