1. Absolute max and min

Find the absolute max and min value(s) for :

a) $f(x) = -x^2 + 4x + 7$

$f'(x) = -2x + 4$

$x = 2$
$y = 11$

critical point $(2,11)$

No absolute min or max value

b) $f(x) = -x^{4} + 4x^{3}$

$f'(x) = -4x^{3} + 12x^2$

$x = 0$
$y = 0$

$x = 3$
$y = 27$

critical points: $(0,0)$ and $(3,27)$

Absolute min at $(0,0)$
Absolute max at $(3,27)$

c)
$f(x) = x + x^{-1} , x > 0$

$f'(x) = 1 - \frac{1}{x}$
$0 = \frac {x - 1}{x}$

$x = 1$
$y = 2$

critical point $(1,2)$

since $x > 0$, there is no absolute max
absolute min value is $(1,2)$

2. Originally Posted by Macleef
Find the absolute max and min value(s) for :

a) $f(x) = -x^2 + 4x + 7$

$f'(x) = -2x + 4$

$x = 2$
$y = 11$

critical point $(2,11)$

No absolute min or max value

b) $f(x) = -x^{4} + 4x^{3}$

$f'(x) = -4x^{3} + 12x^2$

$x = 0$
$y = 0$

$x = 3$
$y = 27$

critical points: $(0,0)$ and $(3,27)$

Absolute min at $(0,0)$
Absolute max at $(3,27)$

c) $f(x) = x + x^{-1} , x > 0$

$f'(x) = 1 - \frac{1}{x}$
$0 = \frac {x - 1}{x}$

$x = 1$
$y = 2$

critical point $(1,2)$

since $x > 0$, there is no absolute max
absolute min value is $(1,2)$
How is x=2 not a max?

since $f(x)=-x^2+4x+7\Rightarrow{f'(x)=-2x+4}\Rightarrow{f''(x)=-2}$

you found the critical point and since $f''(x)<0,\forall{x}\in\mathbb{R}$ we have that $(2,11)$ is a relative maximum

3. the critical point for a) isn't the absolute max or min ... there's only one coordinate point so I just assumed it doesn't mean anything?

4. Originally Posted by Macleef
the critical point for a) isn't the absolute max or min ... there's only one coordinate point so I just assumed it doesn't mean anything?
Oh it means something alright...to find absolute extrema you find all the relative extrema...

Now you test all the x valuse for relative extrema and all the x values of the endpoints of your interval and the value that when put in f(x) yields the larges result is the absolute maximum point and the lowest is obviously the absolute minimum point