logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

Find the absolute maximum and absolute minimum values of $f$ on the given interval: $\;f(x)=\sqrt{9-x^{2}} ,[-1 , 2]$

Note: This is part 4th of a 7 part question, split as 7 separate questions here.

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • To find the absolute maximum and minimum values of a continuous function f on a closed interval $[a,b]$
  • (i) Find the values of the critical numbers of f in $(a,b)$
  • (ii) Find the value of $f(a)$ and $f(b)$
  • (iii) The largest of the values from (i) and (ii) is the absolute maximun value, the smallest of these values is the absolute minimum value.
$f(x)=\sqrt {9-x^2}$ is continous in $[-1,2]$
Step 1:
$f'(x)= \large\frac{-2x}{2 \sqrt {9-x^2}}=\large\frac{-x}{\sqrt {9-x^2}}$ is defined in $[-1,2]$
$f'(x)=0=>\large\frac{-x}{\sqrt {9-x^2}}$$=0$
Step 2:
$x=0$ is a critical number
The critical value is $f(0)=0$
Now the values of $f$ at the end points are
$f(-1)=\sqrt {9-1}=2 \sqrt 2$
$f(2)=\sqrt {9-4}=\sqrt 5$
Step 3:
Comparing the three values,
The absolute minimum is $f(-1)=0$
and the absolute maximum is $f(2)=\sqrt 5$
answered Jul 31, 2013 by meena.p
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...