Page:Calculus Made Easy.pdf/127

(5) Find the maxima and minima of the function

$y={\dfrac {x^{2}-5}{2x-4}}$.

We have

${\dfrac {dy}{dx}}={\dfrac {(2x-4)\times {2x}-(x^{2}-5)2}{(2x-4)^{2}}}=0$.

for maximum or minimum; or

${\dfrac {2x^{2}-8x+10}{(2x-4)^{2}}}=0$;

or $x^{2}-4x+5=0$; which has for solutions

$x={\tfrac {5}{2}}\pm {\sqrt {-1}}$.

These being imaginary, there is no real value of $x$ for which ${\dfrac {dy}{dx}}=0$; hence there is neither maximum nor minimum.

(6) Find the maxima and minima of the function

$(y-x^{2})^{2}=x^{5}$.

This may be written $y=x^{2}\pm {x^{\frac {5}{2}}}$.

${\dfrac {dy}{dx}}=2x\pm {\tfrac {5}{2}}x^{\frac {3}{2}}=0$ for maximum or minimum;

that is, $x(2\pm {\tfrac {5}{2}}x^{\frac {1}{2}})=0$, which is satisfied for $x=0$, and for $2\pm {\tfrac {5}{2}}x^{\tfrac {1}{2}}=0$, that is for $x={\tfrac {16}{25}}$. So there are two solutions.

Taking first $x=0$. if $x=-0.5$, $y=0.25\pm {\sqrt[{2}]{-(.5)^{5}}}$, and if $x=+0.5$, $y=0.25\pm {\sqrt[{2}]{.55}}$. On one side $y$ is imaginary; that is, there is no value of $y$ that can be represented by a graph; the latter is therefore entirely on the right side of the axis of $y$ (see Fig. 30).

On plotting the graph it will be found that the