Question

Consider the function $f\left(x\right)=x^2-4x+17.$ If $M$ and $m$ are the maximum and minimum values of $f$ in $[0,3]$ respectively, then the value of $2M-m$ is

Answer 1

$f\left(x\right)=x^2-4x+17$
Here, $a=1>0$
$D=-52<0$
So, the graph of $y=f\left(x\right)$ lies above $x$-axis.
The graph of $y=f\left(x\right)$ looks like


Minimum value occurs at $x=\frac{-b}{2a}=2$
$\therefore m=f\left(2\right)=13$
Maximum value of $f$ will occur at one of the end points of the interval $[0,3]$
$f\left(0\right)=17$ and $f\left(3\right)=14$
$\therefore M=max\left\{f\right(0),f(3\left)\right\}=17$
Hence, $2M-m=21$