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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: $ (ii)\: g (x) = x^3-3x$

This is second part of multipart q3

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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:
$g(x)=x^3-3x$
Let $g(x)=x^3-3x$
Differentiating with respect to $x$
$g'(x)=3x^2-3$
$\quad\;\;\;\;=3[x^2-1]$
$\quad\;\;\;\;=3(x+1)(x-1)$
Now $g'(x)=0$
$\Rightarrow 3(x-1)(x+1)=0$
$\Rightarrow$ either $x=1$ or $x=-1$
Step 2:
At $\;x=1$
When $x$ is slightly <1
$g'(x)$ is -ve
When $x$ is slightly >1
$g'(x)$ is +ve
Thus $g'(x)$ changes sign from -ve to +ve as $x$ increases through 1 and hence $x=1$ is a point of local minimum.
Step 3:
At $\;x=-1$
When $x$ is slightly <-1
$g'(x)$ is +ve
When $x$ is slightly >-1
$g'(x)$ is -ve
Thus $g'(x)$ changes sign from positive to negative as $x$ increases through -1 and hence $x=-1$ is a point of local maximum.
Local minimum value =$f(1)=1-3=-2$
Local maximum value =$f(-1)=-1+3=2$
answered Aug 6, 2013 by sreemathi.v
edited Aug 19, 2013 by sharmaaparna1
 

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