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.

$$g(x) = (b + c)x^2 + bx + c$$
$$g'(x) = 2(b + c)x + b$$
$$g''(x) = 2(b + c)$$
For $f$ to be continuous at $k$, we must have $g(k) = g'(k)$,
$$(b + c)k^2 + bk + c = 2(b + c)k + b$$
$$(b + c)k^2 + (- b - 2c)k + (c - b) = 0$$
$$k = \frac{b + 2c \pm \sqrt{(-b-2c)^2 - 4(b+c)(c-b)}}{2(b+c)}$$
$$k = \frac{b + 2c \pm \sqrt{5b^2+4bc}}{2(b+c)}$$
For $f$ to be differentiable at $k$, we must have $g'(k) = g''(k)$.
$$2(b + c)k + b = 2(b + c)$$
$$2(b + c)k = b + 2c$$
$$k = \frac{b + 2c}{2(b + c)}$$
Equating the two expressions for $k$ gives:
$$\frac{b + 2c \pm \sqrt{5b^2+4bc}}{2(b+c)} = \frac{b + 2c}{2(b + c)}$$
$$b + 2c \pm \sqrt{5b^2+4bc} = b + 2c$$
$$\pm \sqrt{5b^2+4bc} = 0$$
$$5b^2+4bc = 0$$
$$b(5b+4c) = 0$$
$$b = 0 \text{ or } b = -\frac{4c}{5}$$
If $b = 0$, then $k = \frac{2c}{2c} = 1$.
If $b = -\frac{4c}{5}$, then $k = \frac{-\frac{4c}{5} + 2c}{2(-\frac{4c}{5} + c)} = \frac{1.2c}{0.4c} = 3$.
So $k \in \{ 1, 3 \}$.  The maximum value in this set is 3.
