Page:Calculus Made Easy.pdf/115

Plot these values as in Fig. 27.

It will be evident that there will be a maximum somewhere between $x=1$ and $x=2$; and the thing looks as if the maximum value of $y$ ought to be about $x=2{\tfrac {1}{4}}$. Try some intermediate values. If $x=1{\tfrac {1}{4}}$, $y=2.187$; if $x=1{\tfrac {1}{2}}$, $y=2.25$; if $x=1.6$, $y=2.24$. How can we be sure that $2.25$ is the real maximum, or that it occurs exactly when $x=1{\tfrac {1}{2}}$?

Now it may sound like juggling to be assured that there is a way by which one can arrive straight at a maximum (or minimum) value without making a lot of preliminary trials or guesses. And that way depends on differentiating. Look back to an earlier page (81) for the remarks about Fig. 14 and Fig. 15, and you will see that whenever a curve gets either to its maximum or to its minimum height, at that point its ${\dfrac {dy}{dx}}=0$. Now this gives us the clue to the dodge that is wanted. When there is put before you an equation, and you want to find that value of $x$ that will make its $y$ a minimum (or a maximum), first differentiate it, and having done so, write its ${\dfrac {dy}{dx}}$ as equal to zero, and then solve for $x$. Put this particular value of $x$ into the original equation, and you will then get the required value of $y$. This process is commonly called “equating to zero.”

To see how simply it works, take the example with which this chapter opens, namely

$y=x^{2}-4x+7$.