Values that make a piecewise function differentiable This is a common type of question in calculus courses, but I have found the reasoning in the answer given lacking. What values of $b$ make $f(x)$ differentiable for all x.
$$
f(x) = \left\{
        \begin{array}{ll}
            bx^2-3 & \quad x \leq -1 \\
            3x+b & \quad x > -1
        \end{array}
    \right.
$$
The standard way given in the answer was to first calculate $f'(x)$ by differentiating both pieces. (We only need to check the meeting point of since both pieces are clearly differentiable)
$$f'(x) = \left\{
        \begin{array}{ll}
            2bx & \quad x < -1 \\
            3 & \quad x > -1
        \end{array}
    \right.
$$
1) Then, finding the values that make the left and right limits of the derivative equal $\lim_{x\to-1^+}f'(x)=\lim_{x\to-1^-}f'(x)$ when $ b=-\frac{3}{2}$. 
2) If you cannot make them equal, then you say that the function is not differentiable at $x=-1$
The problem I see with 1) and 2) is that 
1) $\lim_{x\to c^+}f'(x)=\lim_{x\to c^-}f'(x)$ does not imply that $f'(c)$ exists.For example $$f(x) = \left\{
        \begin{array}{ll}
            x^2 & \quad x < -1 \\
            x^2+1 & \quad x > -1
        \end{array}
    \right.
$$
2) $\lim_{x\to c^+}f'(x)$ or $\lim_{x\to c^-}f'(x)$ not existing, therefore not equal, does not imply that $f'(c)$ does not exist. For example, $$f(x) = \left\{
        \begin{array}{ll}
            x^2\cos(\frac{1}{x}) & \quad x \ne0 \\
            0 & \quad x =0
        \end{array}
    \right.
$$
It seems to me that the approach given by the answer is not correct and we have to always check for the definition at the point. 
 A: In problems like these in a calculus course you typically have
$$
f(x) = \begin{cases}
g(x), & x \le a, \\
h(x), & x > a,
\end{cases}
$$
where $g(x)$ and $h(x)$ are very nice functions (differentiable on the whole real line, for example). Let's say that they are differentiable at $a$, at least.
Then, what you can say right away (just from the definitions) is that the derivative from the left is
$$
f'_{-}(a) = g'(a)
$$
and that IF $g(a)=h(a)$ so that $f$ is continuous at $a$, then also the derivative from the right is
$$
f'_{+}(a) = h'(a)
.
$$
This does not involve taking limits of $g'(x)$ or $h'(x)$, it's just their derivatives at the point $a$ that's involved. And those derivatives should be computed in the proper way; just use differentiation rules in simple cases like if $g$ and $h$ are polynomials or such, but use the definition in more complicated cases like the $x^2 \cos(1/x)$ example.
However, if $g(a) \neq h(a)$, then the right-hand derivative $f'_{+}(a)$ does not exist!
So you really need to check the continuity as well. In your example, the value of $b$ that you found just happens to make the function continuous too, but that was a fluke. What you should have done first is to find $b$ to make the function continuous, and then check if the right and left derivatives agree for that value of $b$. (Not the limits of $f'(x)$ from the right/left, but the actual right/left derivatives. That happens to be the same in this case, but in principle it need not be, as you yourself pointed out.)
A: If $\lim_{x\to c+} f'(x)=\lim_{x\to c-} f'(x)$ we cannot say that $f'(c)$ exists. [This condition does not involve the value of $f$ at $c$ so it cannot guarantee existence of the derivative]. What is true is $f'(c)$ exists iff $\lim_{x\to c+} \frac {f(x)-f(c)} {x-c}=\lim_{x\to c-} \frac{f(x)-f(c)} {x-c}$.
