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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: $ f (x) =x^3-6x^2+9x+15$

$\begin{array}{1 1}Minimum = 15\; Maximum = 20 \\ Minimum = 17\; Maximum = 19 \\ Minimum = 15\; Maximum = 19 \\ Minimum = 15 \;Maximum = 18 \end{array} $

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Toolbox:
  • $f$ 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:
Let $f(x)=x^3-6x^2+9x+15$
Differentiating with respect to x
$f'(x)=3x^2-12x+9$
$\qquad=3[x^2-4x+3]$
$\qquad=3(x-1)(x-3)$
For Maxima and Minima
$f'(x)=0$
$\Rightarrow 3(x-1)(x-3)=0$
$x=1,3$
Step 2:
Now At $x=1$
When $x$ is slightly < 1$f'(x)=(-)(-)=+ve$
When $x$ is slightly > 1$f'(x)=(+)(-)=-ve$
$f'(x)$ changes sign from $+ve$ to $-ve$ as $x$ increases through 1.
$\Rightarrow f(x)$ has a local maximum at $x=1$
Local maximum value =$f(1)=x^3-6x^2+9x+15$
$\qquad\qquad\qquad\qquad=(1)^3-6(1)+9(1)+15$
$\qquad\qquad\qquad\qquad=1-6+9+15$
$\qquad\qquad\qquad\qquad=19$
Step 3:
At $x=3$
When $x$ is slightly < 3$f'(x)=(+)(-)=-ve$
When $x$ is slightly > 3$f'(x)=(+)(+)=+ve$
$\therefore f'(x)$ changes sign from -ve to +ve as $x$ increases through 3.
$\Rightarrow f(x)$ has a local minimum at $x=3$
$\therefore$ Local minimum value $f(3)=x^3-6x^2+9x+15$
$\qquad\qquad\qquad\qquad\qquad=(3)^3-6(3)^2+9(3)+15$
$\qquad\qquad\qquad\qquad\qquad=27-54+27+15$
$\qquad\qquad\qquad\qquad\qquad=15$
Step 4:
Method to determine the change of sign of $f(x)$ as $x$ increase through a particular point.
Let $f'(x)=(x-a)(x-b)(x-c)$
At $x=a$
$(x-a)=a-a=0$
$(x-b)=a-b$
$(x-c)=a-c$
Sign corresponding to $x-a$ changes from -ve to +ve sign corresponding to $x-b$ is that of $a-b$ and remains to be the same.
Sign corresponding to $x-c$ is that of $a-c$ and remains to be the same.
Step 5:
Let $f'(x)=(x-1)(x-3)$
At $x=1$
$x-1=1-1=0\rightarrow -ve$
$x-3=1-3=-2\rightarrow -ve$
Factor $(x-1)(x-3)$
When $x$ is slightly $<1(-)(-)\Rightarrow +ve$
When $x$ is slightly $>1(+)(-)\Rightarrow -ve$
$\therefore f'(x)$ changes from +ve to -ve.
Step 6:
At $x=3$
$x-1=3-1=2=+ve$
$x-3=3-3=0$
Factor $(x-1)(x-3)$
When $x$ is slightly $<3(+)(-)\Rightarrow -ve$
When $x$ is slightly $>3(+)(+)\Rightarrow +ve$
$\therefore$ f'(x) changes sign from $-ve$ to $+ve$
answered Aug 7, 2013 by sreemathi.v
edited Aug 19, 2013 by sharmaaparna1
 

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