logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
0 votes

The least value of the function $f(x)=ax+\frac{b}{x}(a>0,b>0,x>0)$ is ___________.

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • To obtain the absolute maxima or minima for the function $f(x)$
  • (i) Find $f'(x)$ and put $f'(x)=0$
  • (ii) Obtain the points from $f'(x)=0$
  • (iii) By phthogoras theorem, find $f''(x)$ and check the value of $f''(x)$ for each of the points obtained if $f''(x)>0$ it has a minimum at that point. If $f''(x)<0$ then it has a maximum at that point.
Step 1
$ f(x)=ax+ \large\frac{b}{x}$
differentiating w.r.t $x$ we get,
$ f'(x)=a-\large\frac{b}{x^2}$
when $f'(x)=0$
$ \Rightarrow a-\large\frac{b}{x^2}=0$
$ \Rightarrow a = \large\frac{b}{x^2}$
$ x^2=\large\frac{b}{a}$
$ x = \sqrt{\large\frac{b}{a}}$
differentiating $f'(x)$ w.r.t $x$ we get,
$ \therefore f''(x)= \large\frac{b}{x^3}$
Hence the given function has minimum value.
Substituting the value of $x$ in $y$ we get,
$ y=a \sqrt{\large\frac{b}{a}}+b\sqrt{\large\frac{a}{b}}$
$=2\sqrt{ab}$
Hence the least value of the given function is $2\sqrt {ab}$
answered Aug 13, 2013 by thanvigandhi_1
 

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
...