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## 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 $A$
From ($i$), the minimum value of $A$ is :
 ${\left. {{A_{\min }} = \dfrac{{ - 1}}{{2m}}{{\left( {k - mh} \right)}^2}} \right|_{m = \dfrac{{ - k}}{h}}} = 2hk$