Find the absolute extrema of the function $f(x)=x^2-2x-2$ on $[0,1]$
To find a the extrema of a continuous function $f$ on a closed interval $[a,b]$, use the following steps:
1) Find the critical numbers of $f$ on the open interval $(a,b)$.
2) Evaluate $f$ at each of its critical numbers in $(a,b)$
3) Evaluate $f$ at the endpoints, $a$ and $b$
4) The least of these values is the absolute minimum, the greatest of these is the maximum.
Solution:
Since our function is continuous on $[0,1]$, we can just follow the above steps to find the absolute maximum and absolute minimum.
Let's find the critical numbers by solving $f'(x)=0$:
$f'(x)=2x-2=0$
$\rightarrow x=1$
And so all the points where absolute extrema could happen are when $x=1$ and then the boundary points, which are $x=0$ and $x=1$. So in this case, the value of $x$ such that $f'(x)=0$ was already one of our boundary points! So the values we need to evalute are $f(0)$ and $f(1)$:
$f(0)= 0^2 -2(0)-2=-2$
$f(1)= 1^2 -2(1)-2 =-3$
So our absolute maximum is $(0,-2)$ and our absolute minimum is $(1,-3)$
$\endgroup$ 21 Answer
$\begingroup$No calculus is needed if you remember how to complete the square: $f(x)=x^2-2x-2=(x-1)^2-3$.
Thus the minimum is at $x=1$ and the max at $x=0$.
$\endgroup$
