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100

Calculus Made Easy

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.

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