USCTrojanTennis' question at Yahoo Answers regarding optimization

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SUMMARY

The discussion addresses the relationship between the extrema of a function \( f(x) \) and its square \( f^2(x) \). It establishes that if \( f(x) > 0 \), the critical points of \( f(x) \) coincide with those of \( f^2(x) \) due to the derivative \( y'(x) = 2f(x)f'(x) \). The second derivative test confirms that the nature of the extrema remains consistent between \( f(x) \) and \( f^2(x) \). This conclusion is reached through differentiation and the application of the chain and product rules.

PREREQUISITES
  • Understanding of calculus, specifically differentiation
  • Familiarity with the chain rule and product rule
  • Knowledge of critical points and extrema
  • Basic concepts of functions and their properties
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  • Study the implications of the second derivative test in calculus
  • Explore the properties of even and odd functions in relation to extrema
  • Learn about optimization techniques in calculus
  • Investigate the behavior of functions under transformations, such as squaring
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Students and educators in mathematics, particularly those focusing on calculus and optimization techniques, will benefit from this discussion. It is also valuable for anyone looking to deepen their understanding of function behavior and critical point analysis.

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Hello USCTrojanTennis,

If we look at:

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

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

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

Since we are given $$f(x)\ne0$$, 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:

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

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

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

Since we are given $$0<f(x)$$, 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 http://www.mathhelpboards.com/f10/ forum.

Best Regards,

Mark.
 

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