<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] Equating $g(x)$ and $g'(x)$ and using $c=a-b$ gives
$$ax^2+x(b-2a)=2b-a.$$
Completing the square:
$$(2ax+b-2a)^2=4a(2b-a)+(b-2a)^2=b^2+4ab.$$
We must have only one solution, that is $b=0$ or $b=-4a$.  From $(2ax+b-2a)^2=0$ we get $x = 1$ or $x = 3$.
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