What are the absolute extrema of f(x)=2x^2 - 8x + 6 in[0,4]?

Mar 9, 2016

$6$ and $- 2$

Explanation:

Absolute extrema (the min. and max. values of a function over an interval) can be found by evaluating the endpoints of the interval and the points where the derivative of the function equals 0.

We begin by evaluating the endpoints of the interval; in our case, that means finding $f \left(0\right)$ and $f \left(4\right)$:
$f \left(0\right) = 2 {\left(0\right)}^{2} - 8 \left(0\right) + 6 = 6$
$f \left(4\right) = 2 {\left(4\right)}^{2} - 8 \left(4\right) + 6 = 6$
Note that $f \left(0\right) = f \left(4\right) = 6$.

Next, find the derivative:
$f ' \left(x\right) = 4 x - 8 \to$using the power rule
And find the critical points; i.e. the values for which $f ' \left(x\right) = 0$:
$0 = 4 x - 8$
$x = 2$
Evaluate the critical points (we only have one, $x = 2$):
$f \left(2\right) = 2 {\left(2\right)}^{2} - 8 \left(2\right) + 6 = - 2$

Finally, determine the extrema. We see that we have a maximum at $f \left(x\right) = 6$ and a minimum at $f \left(x\right) = - 2$; and since the question is asking what the absolute extrema are, we report $6$ and $- 2$. If the question was asking where the extrema occur, we would report $x = 0$, $x = 2$, and $x = 4$.