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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the maximum and minimum values, if any, of the following functions given by $ (v) \: h (x) =x+1,x\in (-1,1)$

This is fifth part of multipart q2

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  • $f$ is said to have a maximum value in $I$ , if there exist a point c in I such that $f(c) \geq f (x)$ for all $x \in I$.The number $f( c)$ is called the maximum value of f in I and the point c is called a point of maximum value of f in I
  • $f$ is said to have a minimum value in $I$ , if there exist a point $c$ in I such that $f(c) \leq f (x)$ for all $x \in I$.The number $f(c)$ is called the minimum value of f in I and the point $c$ in this case is called a point of minimum value of $f$ in I
  • $f$ is said to have a extreme value in $I$ , if there exist a point $c$ in I such that f(c) is either a maximum value or minimum value of $f$ in $I$. The number $f (c)$ in this case is called the extreme value of $f$ in $I$ and the point $c$ is called the extreme point.
Step 1:
$h(x)=x+1,x\in (-1,1)$
Let $f(x)=x+1$
As $x\rightarrow \infty f(x)\rightarrow \infty$
Also $x\rightarrow -\infty f(x)\rightarrow -\infty$
Thus there is no maximum or minimum value.
Step 2:
However at $x=1 \;f(x)=1+1=2$
At $x=-1$ f(x)=-1+1=0$
Therefore greatest value of $f(x)$ is 2 and least value is 0.
answered Aug 6, 2013 by sreemathi.v
 

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