ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 16

Sunday, 3 August 2014

CHAPTER 16 - Worked Out Examples

Example: 1

Find the extrema points of $f\left( x \right) = 3{x^4} - 4{x^3} - 36{x^2} + 28$ .

Solution: 1

	$f'\left( x \right) = 12{x^3} - 12{x^2} - 72x$
	$= 12x\left( {{x^2} - x + 6} \right)$
	$= 12x\left( {{x^2} - x + 6} \right)$

We determine the sign of $f'(x)$ using a number line:

From the number line, observe that (using the $FODT$ ):

	$x = - 2\,{\rm{and }}\,x = 3$ are local minima
	$x = 0$ is a local maximum

Alternatively, we can use the $SODT$ :

	$f''\left( x \right) = 36{x^2} - 24x - 72$
	$= 12\left( {3{x^2} - 2x - 6} \right)$

	$f''\left( 0 \right) < 0 \,\,\,\, \Rightarrow \,\,\,\, x = 0$ is a local maximum
	$f''\left( { - 2} \right) > 0 \,\,\,\, \Rightarrow \,\,\,\, x = - 2$ is a local minimum
	$f''\left( 3 \right) > 0 \,\,\,\, \Rightarrow \,\,\,\, x = 3$ is a local minimum

Example: 2

Let $f\left( x \right) = 2{x^3} - 3\left( {a + b} \right){x^2} + 6abx.$ If $a < b$ , determine the local maximum/minimum points of $f(x)$ . If $a = b$ , how will the answer change?

Solution: 2

	$f'(x) = 6{x^2} - 6(a + b)x + 6ab$
	$= 6(x - a)(x - b)$

To determine the sign of $f'(x)$ in different intervals, we use a number line:

	$f'(x) > 0\,\,:\,\,x < a$ or $x>b$
	$f'(x) < 0\,\,:\,\,a < x < b$

Observe that $f'(x)$ changes from positive to negative in the neighbourhood of $x = a$ .

$\Rightarrow\,\,\,\, x = a$ is a point of local maximum

Similarly, $f'(x)$ changes from negative to positive in the neighbourhood of $x = b$ .

$\Rightarrow\,\,\,\, x = b$ is a point of local minimum

If $a = b,f'(x) = 6{(x - a)^2}$

Notice that $f'(x)$ is never negative. $f'(x)$ is always positive except at $x = a$ where $f'(x) = 0$

$\Rightarrow\,\,\,\, x = a$ is a point of inflexion

Example: 3

Let $(h,k)$ be a fixed point, where $h > 0,k > 0$ . A straight line passing through this point cuts the positive direction of the co-ordinate axes at the points $P$ and $Q$ . Find the minimum area of $\Delta OPQ$ , $O$ being the origin.

Solution: 3

The given point $(h,k)$ will lie in the first quadrant.

Convince yourself that there will be a particular slope of $PQ$ at which the area of $\Delta OPQ$ is minimum. If the $\left( {{\rm{slope}}} \right) \to 0$ or $\left( {{\rm{slope}}} \right) \to \infty$ , $\left( {{\rm{area}}} \right) \to \infty$ . However, at some finite slope in between these two extremes, area will assume a minimum value.

Assume $m$ to be the slope we wish to determine so that area is minimum (Notice that $m$ will be negative). We first write down the equation of a straight line passing through $(h,k)$ with slope $m$ :

$y - k = m(x - h)$

This cuts the $X$ -axis at $P\left( {h - \dfrac{k}{m},0} \right)$ and the $Y$ -axis at $Q(0,k - mh)$

Assume $A$ to be the area of $\Delta OPQ$ .

Therefore,

	$A = \dfrac{1}{2} \times OP \times OQ$
	$= \dfrac{1}{2} \times \left( {h - \dfrac{k}{m}} \right) \times \left( {k - mh} \right)$
	$= \dfrac{{ - 1}}{{2m}}{\left( {k - mh} \right)^2}$	$\ldots(i)$

For $A$ to be minimum,

	$\dfrac{{dA}}{{dm}} = 0\,\,\,{\rm{and}}\;\,\dfrac{{{d^2}A}}{{{d^2}m}} > 0$
	$\Rightarrow \,\,\,\, \dfrac{{dA}}{{dm}} = \dfrac{{ - 1}}{{2m}}.2\left( {k - mh} \right).\left( { - h} \right) + {\left( {k - mh} \right)^2}.\dfrac{1}{{2{m^2}}}$
	$= \dfrac{1}{{2{m^2}}}\left( {{k^2} - {m^2}{h^2}} \right)$
	$\Rightarrow \,\,\,\, \dfrac{{dA}}{{dm}} = 0\,\,\,{\rm{when }}{k^2} - {m^2}{h^2} = 0$
	$\Rightarrow \,\,\,\, m = \pm \dfrac{k}{h}$

Since $m$ must be negative, $m = \dfrac{{ - k}}{h}$ .

Now,

	${\left. {\dfrac{{{d^2}A}}{{d{m^2}}}} \right\|_{m = \dfrac{{ - k}}{h}}}{\left. { = \dfrac{1}{{2{m^2}}}\left( { - 2m{h^2}} \right) - \dfrac{{\left( {{k^2} - {m^2}{h^2}} \right)}}{{{m^3}}}} \right\|_{m = \dfrac{{ - k}}{h}}}$
	$= \dfrac{{{h^3}}}{k} > 0$
	$\Rightarrow \,\,\,\,m = \dfrac{{ - k}}{h}$ is a point of local minimum for $A$