Find the numbers A,B such that the derivative function
\begin{cases}
Ax^3+Bx+2& \text{if }x=<2,\\
Bx^2-A & \text{if } x>2
\end{cases}
is everywhere continuous .
My work:
Let's name the function $f(x)$. Derivative every part of $f(x)$ and we get this:
\begin{cases}
3Ax^2+B& \text{if }x=<2,\\
2Bx & \text{if } x>2
\end{cases}
Note $f'$ is continuous everywhere except perhaps 2, since it is a composition/product of continuous functions.
To work out continuous at 2 left and right limit of function at point 2 must be equal.
$$\lim_{x\to 2^+}2Bx=4B $$
and
$$\lim_{x\to 2^-}3Ax^2+B =12A+B$$
We only know that $4B=12A+B \to B=4A$
Is my work ok?
Please consider limiting yourself to one question per post.
For the first problem, note that if we want $f'$ to be continuous at $x = 2$, then $f'$ first needs to be well-defined at $x = 2$. So we know that $f$ must be differentiable at $x = 2$, implying that $f$ must be continuous at $x = 2$. So we know that $\lim_{x \to 2} f(x)$ must exist, implying that: \begin{align*} \lim_{x \to 2^-} f(x) &= \lim_{x \to 2^+} f(x) \\ A(2)^3 + B(2) + 2 &= B(2)^2 - A \\ 9A - 2B &= -2 \end{align*} Combine this with the other constraint to obtain a system of equations that can be solved for $A$ and $B$.