USCTrojanTennis' question at Yahoo! Answers regarding optimization

Staff member

MarkFL

Staff member
Hello USCTrojanTennis,

If we look at:

$$\displaystyle y(x)=f^2(x)$$

we find by differentiating with respect to $x$, using the chain rule:

$$\displaystyle y'(x)=2f(x)f'(x)$$

Since we are given $$\displaystyle f(x)\ne0$$, we know that the roots of the derivative are simply those from $f'(x)$, thus we get the same critical numbers. To demonstrate that the nature of the extrema are the same in both cases, we may differentiate again, this time using the product rule:

$$\displaystyle y''(x)=2\left(f(x)f''(x)+f'^2(x) \right)$$

Now, at the critical points $x=c$, we of course have $f'(c)=0$, hence:

$$\displaystyle y''(c)=2f(c)f''(c)$$

Since we are given $$\displaystyle 0<f(x)$$, then at the critical points the sign of the second derivative of $y$ is the same as the sign of the second derivative of $f$, and so we have shown that the function $f(x)$ and its square $f^2(x)$ have the same extrema.

To USCTrojanTennis and any other guests viewing this topic, I invite and encourage you to register and post other calculus problems in our Calculus forum.

Best Regards,

Mark.