Question
Determine the values of $a$ and $b$ such that the following function is differentiable at 0.
$$f(x) =
\begin{cases}
ax^3cos(\frac 1 x) + bx + b, & \text{if }x \lt 0 \\
\sqrt{a + bx}, & \text{if }x \geq 0
\end{cases}$$
My solution
For $f$ to be differentiable at $0$, $f$ must first be continuous at $0$.
$$\implies \lim\limits_{x\to0^-}f(x) = \lim\limits_{x\to0^+}f(x)$$
Consider
\begin{align}
\lim\limits_{x\to0^-}f(x) & =
\lim\limits_{x\to0^-}[ax^3cos(\frac 1 x) + bx + b].
\\[5 mm]
\because \lim\limits_{x\to0^-}ax^3cos(\frac 1 x) & =
\lim\limits_{x\to0^-}bx
\\[5 mm] & =
0
\end{align}
$$\therefore \lim\limits_{x\to0^-}f(x) = b$$
Then, consider
\begin{align}
\lim\limits_{x\to0^+}f(x) & =
\lim\limits_{x\to0^+}\sqrt{a + bx}
\\[5 mm] & =
\sqrt{a}.
\end{align}
$$\implies a = b^2$$
Furthermore, for $f$ to be differentiable at $0$,
$$\lim\limits_{x\to0^-}\frac {f(x) – \sqrt{a}} x = \lim\limits_{x\to0^+}\frac {f(x) – \sqrt{a}} x$$
When $a = b^2$,
\begin{align}
\lim\limits_{x\to0^+}\frac {f(x) – \sqrt{a}} x & =
\lim\limits_{x\to0^+}\frac {\sqrt{a + bx} – \sqrt{a}} x
\\[5 mm] & =
\lim\limits_{x\to0^+}\frac b {\sqrt{a + bx} + \sqrt{a}}
\\[5 mm] & =
\frac b {2\sqrt{a}}
\\[5 mm] & =
\frac 1 2
\\[5 mm]
\implies \lim\limits_{x\to0^-}\frac {f(x) – \sqrt{a}} x & =
\lim\limits_{x\to0^-}\frac {ax^3cos(\frac 1 x) + bx + b – \sqrt{a}} x
\\[5 mm] & =
\lim\limits_{x\to0^-}\frac {b^2x^3cos(\frac 1 x) + bx} x
\\[5 mm] & =
\lim\limits_{x\to0^-}[b^2x^2cos(\frac 1 x) + b]
\\[5 mm] & =
\frac 1 2
\end{align}
$$\because \lim\limits_{x\to0^-}b^2x^2cos(\frac 1 x) = 0$$
$$\therefore b = \frac 1 2$$
$$\implies a = \frac 1 4$$
I would like to know if my proposed solution is logical and correct. Moreover, any alternative solutions that are more elegant or succinct are welcomed as well 🙂
Thank you all in advance!
Edit
Following a discussion with MPW, looks like all is well, except perhaps the fact that
$$\implies a = b^2$$
should have been left as
$$\implies \sqrt{a} = b$$
Best Answer
It looks essentially correct and nicely detailed to me, good work.
My only comment would be that you're assuming $b\geq 0$ when you use $\sqrt{b^2} = b$, so it might be worth considering the computations under the assumption $b<0$ to see if you get a second solution (I haven't pursued this). Note this would give that $\sqrt{b^2} = -b$.
On the face of it, I don't see an immediate reason to exclude $b<0$.