The denominator of a fraction is 16 more than | vedclass

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(B) Let the numerator be $x$. Then the denominator is $x^2 + 16$.The fraction is $f(x) = \frac{x}{x^2 + 16}$.To find the extrema,we find the derivativ

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$ - 1/8$

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(B) Let the numerator be $x$. Then the denominator is $x^2 + 16$.The fraction is $f(x) = \frac{x}{x^2 + 16}$.To find the extrema,we find the derivative $f'(x)$:$f'(x) = \frac{(x^2 + 16)(1) - x(2x)}{(x^2 + 16)^2} = \frac{16 - x^2}{(x^2 + 16)^2}$.Setting $f'(x) = 0$ gives $16 - x^2 = 0$,so $x = 4$ or $x = -4$.We evaluate $f(x)$ at these critical points:For $x = 4$,$f(4) = \frac{4}{16 + 16} = \frac{4}{32} = \frac{1}{8}$.For $x = -4$,$f(-4) = \frac{-4}{16 + 16} = \frac{-4}{32} = -\frac{1}{8}$.Comparing the values,the least value is $-1/8$.

The denominator of a fraction is $16$ more than the square of the numerator. Find the least value of the fraction.

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