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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 $ f (x) = -\mid x+1\mid+3$

$\begin{array}{1 1}Maximum = 0 \\ Maximum= 4 \;Minimum =0 \\ Maximum=3,No\;minimum \;value \\ Maximum =8\;Minimum =0 \end{array} $

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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.
Let $f(x)=-\mid x+1\mid+3$
Now $\mid x+1\mid\geq 0$ for all $x\in R$
$\Rightarrow -\mid x+1\mid+3\leq 0$ for all $x\in R$
$\therefore -\mid x+1\mid+3\leq 3\Rightarrow f(x)\leq 3$
Thus the maximum value of $f(x)$ is 3 and there is no minimum value.
answered Aug 6, 2013 by sreemathi.v
 

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