Page:Calculus Made Easy.pdf/126

(3) Find the maxima and minima of the function

$y={\dfrac {x}{4-x}}+{\dfrac {4-x}{x}}$.

We get

${\dfrac {dy}{dx}}={\dfrac {(4-x)-(-x)}{(4-x)^{2}}}+{\dfrac {-x-(4-x)}{x^{2}}}=0$

for maximum or minimum; or

${\dfrac {4}{(4-x)^{2}}}-{\dfrac {4}{x^{2}}}=0$ and $x=2$.

There is only one value, hence only one maximum or minimum.

${\begin{aligned}{\text{For}}\quad x&=2,{\phantom {.5}}\quad y=2,\\{\text{for}}\quad x&=1.5,\quad y=2.27,\\{\text{for}}\quad x&=2.5,\quad y=2.27;\end{aligned}}$

it is therefore a minimum. (It is instructive to plot the graph of the function.)

(4) Find the maxima and minima of the function $y={\sqrt {1+x}}+{\sqrt {1-x}}$. (It will be found instructive to plot the graph.)

Differentiating gives at once (see example No. 1, p.68)

${\dfrac {dy}{dx}}={\dfrac {1}{2{\sqrt {1+x}}}}-{\dfrac {1}{2{\sqrt {1-x}}}}=0$.

for maximum or minimum.

Hence ${\sqrt {1+x}}={\sqrt {1-x}}$ and $x=0$, the only solution

For $x=0$, $y=0$.

For $x=\pm 5$, $y=1.932$, so this is a maximum.