Show that the absolute minimum value off(x) = x2 + x2(8 x)2 , x> 8occurs at x = 10 by

Chapter 4, Problem 51

(choose chapter or problem)

Show that the absolute minimum value of

$f(x)=x^{2}+\frac{x^{2}}{(8-x)^{2}}, x>8$

occurs at $x=10$ by using a CAS to find $f^{\prime}(x)$ and to solve the equation $f^{\prime}(x)=0$.

Equation Transcription:

$f(x)={x}^{2}+\frac{{x}^{2}}{(8-x{)}^{2}},x>8$

$x=10$

$f'(x)$

$f'(x)=0$

Text Transcription:

f(x)=x^2+x^2/(8-x)^2,x>8

x=10

f'(x)

f'(x)=0

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