Page:Calculus Made Easy.pdf/127

 (5) Find the maxima and minima of the function

We have

for maximum or minimum; or

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

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

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

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 $$x=+0.5$$, $y=0.25\pm{\sqrt[2]{.55}}$|$y=0.25\pm{\sqrt[2]{(.5)^5}}$|undefined [sic]. 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