Question

If $f\left(x\right)=\{\begin{array}{cc}x,& x\le 1\\ x^2+bx+c,& x>1\end{array}$
is a differentiable function, then the value of $5c-8b$ is

Answer 1

$f\left(x\right)=\{\begin{array}{cc}x,& x\le 1\\ x^2+bx+c,& x>1\end{array}$

$f^{\prime }\left(x\right)=\{\begin{array}{cc}1,& x\le 1\\ 2x+b,& x>1\end{array}$

$f$ is a differentiable function.
So, $f$ must be differentiable at $x=1$
Then it must be continuous at $x=1$ also.
$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)=f\left(1\right)\Rightarrow 1=1+b+c\Rightarrow b+c=0$

Also, $\underset{x\to 1^{-}}{lim}{f}^{\prime }\left(x\right)=\underset{x\to 1^{+}}{lim}{f}^{\prime }\left(x\right)$
$\Rightarrow 1=2+b\Rightarrow b=-1\Rightarrow c=1$
$\therefore 5c-8b=13$