Question

Find the absolute maximum value and the absolute minimum value of the given function in the given intervals $f\left(x\right)={(x-1)}^2+3,x\in [-3,-1]$

Answer 1

Given, $f\left(x\right)=(x-1{)}^2+3$

$f^{{}^{\prime }}\left(x\right)=2x-2$

Putting this equal to zero

$f^{{}^{\prime }}\left(x\right)=2x-2=0$

$\therefore x=1$

Now let's look at second derivative of the given function.

$f^{{}^{\prime \prime }}\left(x\right)=2$ and as we can see, this is positive for all $x$

hence $f\left(x\right)$ will have it's minima at $x=1$ but as you can see that $1$ is not in domain of the function so function will have maxima and minima at two extream of domain i.e at $-1$ and $-3$ respectively.

Minimum value of the given function is $f(-1)=(-1-1{)}^2+3=7$

Maximum value of the given function is $f(-3)=(-3-1{)}^2+3=19$