Let $f: \mathbb R \to \mathbb R$, $\begin{eqnarray} f(x)= \begin{cases} x^3+ax^2+b&\quad\text {if } x \le 1 \cr 2x+3&\quad\text{if } x \ge 1\cr \end{cases} \end{eqnarray}$
a) Determine $a$ and $b$ so that $f$ is differentiable at 1.
b) Is $f$ differentiable twice at 1 after determining $a$ and $b$?
My work:
For the continuity, I got $a=4-b$, but for the differential I cant get a decent result when coming from $-$ ($2x+3$ I got to 2, as should)
Hint
Once $2x+3$ is a line, if you want $f$ to be differentiable at $1$ then $2x+3$ must to be tangent to $x^3+ax^2+b$ at $1$, so:
$$f'(x)=(x^3+ax^2+b)'=3x^2+2ax=(2x+3)'=2\\ f'(1)=2=3+2a \rightarrow a=-\frac{1}{2}$$
That give us $b=4-a=\frac{9}{2}$.
Can you finish?
To make it easier, you will realise that you can differentiate $f(x)$ to eliminate $b$.
Determine the derivative $f'(x)$:
$$f'(x)=\begin{cases} 3x^2+2ax&\quad\text {if } x \le 1 \cr 2&\quad\text{if } x \ge 1\cr \end{cases}$$
Thus, you can evaluate limits $x\to 1$ on both sides and set them equal.
$$\lim_{x\to 1^-} (3x^2+2ax)=\lim_{x\to 1^+} (2)$$
$$3+2a=2$$
Thus,
$$\boxed{a=-\frac{1}{2}}$$
Now, you can use this to find $b$ on $f(x)$ so that your function becomes continuous.
$$ f(x)= \begin{cases} x^3-\frac{1}{2}x^2+b&\quad\text {if } x \le 1 \cr 2x+3&\quad\text{if } x \ge 1\cr \end{cases} $$
Evaluate limits:
$$\lim_{x\to 1^-}(x^3-\frac{1}{2}x^2+b)=\lim_{x\to 1^+}(2x+3)$$
$$1-\frac{1}{2}+b=5$$
$$\boxed{b=\frac{9}{2}}$$
For part (b), just differentiate $f(x)$ twice to obtain $f''(x)$ and evaluate limits, and check whether they are equal or not. If you want, you can comment on the answer you get for part (b).