Page:Calculus Made Easy.pdf/120

Test Case.

Let us at once apply our knowledge to a case that we can test.

Let

$y=x^{2}-x$;

and let us find whether this function has a maximum or minimum; and if so, test whether it is a maximum or a minimum.

Differentiating, we get

${\frac {dy}{dx}}=2x-1$.

Equating to zero, we get

$2x-1=0$,

whence

$2x=1$,

or

$x={\tfrac {1}{2}}$.

That is to say, when $x$ is made $={\tfrac {1}{2}}$, the corresponding value of $y$ will be either a maximum or a minimum. Accordingly, putting $x={\tfrac {1}{2}}$ in the original equation, we get

$y=({\tfrac {1}{2}})^{2}-{\tfrac {1}{2}}$,

or

$y=-{\tfrac {1}{4}}$.

Is this a maximum or a minimum? To test it, try putting $x$ a little bigger than ${\tfrac {1}{2}}$,–say make $x=0.6$. Then

$y=(0.6)^{2}-0.6=0.36-0.6=-0.24$,

which is higher up than $-0.25$; showing that $y=-0.25$ is a minimum.

Plot the curve for yourself, and verify the calculation.