<s>[INST] If $g(x)=ax^2+bx+c$ and $f(x)= \begin{cases} {g(x)} & {x\ge k} \\ {g'(x)} & {x< k} \end{cases} $ , $\max (k)=?$ if $f$ is a differentiable function 
If $g(x)=ax^2+bx+c$ and $f(x)= \begin{cases} {g(x)} & {x\ge k} \\
 {g'(x)} & {x< k} \end{cases}  $. If $f(x)$ is a differentiable
function, what is the maximum value of $k$, provided that $b+c=a$?
$1)\frac34\qquad\qquad2)1\qquad\qquad3)3\qquad\qquad4)4$

In order to $f(x)$ be differentiable function, we should have $g(k)=g'(k)=g''(k)$,
$$ak^2+bk+c=2ak+b=2a$$ $$(b+c)k^2+bk+c=(2b+2c)k+b=2b+2c$$
Here for each equation I tried to equate the coefficients of $k^2 , k^1 , k^0$ but I get $a=b=c=0$ which doesn't make sense at all. I don't know how to continue form here.
 [/INST] (A solution with no square roots needed.)
$g'(k) = g''(k)$ gives
$$ 2ak+b = 2a $$
If $a=0$ then this forces $b=0$, $0 = g'(k) = 2a$ so $a=0$, and finally $g(k)=c=0$. In the degenerate case where $f$ and $g$ are zero everywhere, $k$ can be any number at all. The problem should have eliminated this case, to have an answer.
So for the rest, assume $a \neq 0$.
$$ k = 1 - \frac{b}{2a} $$
Then $g(k) = g''(k)$ gives
$$ a\left(1 - \frac{b}{a} + \frac{b^2}{4a^2}\right) + b\left(1-\frac{b}{2a}\right) + c = 2a $$
$$ a + c - \frac{b^2}{4a} = 2a $$
Substituting $c=a-b$,
$$ \frac{b^2}{4a} + b = 0 $$
So either $b=0$ or $\frac{b}{a} = -4$. If $b=0$, $k=1$. If $\frac{b}{a} = -4$, $k=3$. The largest possible value of $k$ (again excluding the $a=b=c=0$ case) is $k=3$.
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