# Thread: absolute maximum /minimum of the function..

1. ## absolute maximum /minimum of the function..

The absolute maximum value of the function

is equal to :

a)None of these mentioned
b)2
c)4
d)0
e)-4

I know that the steps to find absolute minimum or maximus is:

1- find the critical number of f in {a,b}
2-Evaluate f at each critical number in [a,b]
3-Evaluate f at each endpoint of [a,b]
4-The smallest of these values is the minimum and the largest of these values is the maximum.

How can I find the absolute value in the question and the interval is not given ??

2. Whooops, I'm sorry, that was wrong....

Someone please delete this post. Thank you

3. Originally Posted by change_for_better
The absolute maximum value of the function

is equal to :

a)None of these mentioned
b)2
c)4
d)0
e)-4

I know that the steps to find absolute minimum or maximus is:

1- find the critical number of f in {a,b}
2-Evaluate f at each critical number in [a,b]
3-Evaluate f at each endpoint of [a,b]
4-The smallest of these values is the minimum and the largest of these values is the maximum.

How can I find the absolute value in the question and the interval is not given ??
$f(x) = 4 - \sqrt{x}$

domain of $f(x)$ is $x \geq 0$

remember the parent graph $y = \sqrt{x}$ ? you should be able to graph this function using the applicable transformations you learned in precalculus and "see" that the absolute max occurs at the left endpoint, namely $f(0) = 4$.

the calculus confirms this ...

$f'(x) = -\frac{1}{2\sqrt{x}}$

$f'(x) < 0$ for all x in the domain of $f(x)$ except at $x = 0$, where $f'(x)$ is undefined ... $f(x)$ is decreasing throughout its domain, therefore the absolute max is at the left endpoint of the domain.