<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] The original question looks wrong or at least badly worded. From continuity and differentiability at $k$, one indeed gets $g(k)=g'(k)=g''(k)$.
First assume that $a\neq 0$ and therefore $b+c\neq 0$.
Solving the differentiability condition gives $k=1-\frac{b}{2a}=\frac{b+2c}{2b+2c}$, using $b+c=a$.
The continuity condition gives the degree two equation in $k$, $(b+c)k^2 -(b+2c)k+c-b=0$, with discriminant $\Delta=5b^2+4bc$. The solutions are $k_{1,2}=\frac{b+2c\pm \sqrt{5b^2+4bc}}{2b+2c}$.
Comparing the two conditions, assuming $b+c\neq 0$, the only way a $k$ can exist is if $5b^2+4bc=0$, i.e., $b=0$ or $c=-\frac{5b}{4}$.
If $b=0$, then the unique solution is $k=1-\frac{0}{2a}=1$.
If $c=-\frac{5b}{4}$, then again there is a unique solution $k=\frac{b-\frac{5b}{2}}{2(b-\frac{5b}{4})}=3$.
Finally, if $a=b+c=0$, then $g(x)=bx-b$ and the differentiability and continuity condition boils down to $bk-b=b=0$, so $f$ is differentiable if and only if $g\equiv0$. And then clearly any $k$ works.
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