If f(x) = x^2 - x + 5, x frac{1}{2}, and g( | vedclass

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(C) Given $f(x) = x^2 - x + 5$ for $x > \frac{1}{2}$.To find $g'(7)$,we use the formula for the derivative of an inverse function: $g'(y) = \frac{1}{f

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$\frac{1}{3}$

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(C) Given $f(x) = x^2 - x + 5$ for $x > \frac{1}{2}$.To find $g'(7)$,we use the formula for the derivative of an inverse function: $g'(y) = \frac{1}{f'(x)}$,where $y = f(x)$.First,find $x$ such that $f(x) = 7$:$x^2 - x + 5 = 7 \implies x^2 - x - 2 = 0$.$(x - 2)(x + 1) = 0$.Since $x > \frac{1}{2}$,we have $x = 2$.Now,find $f'(x)$:$f'(x) = \frac{d}{dx}(x^2 - x + 5) = 2x - 1$.At $x = 2$,$f'(2) = 2(2) - 1 = 3$.Therefore,$g'(7) = \frac{1}{f'(2)} = \frac{1}{3}$.

If $f(x) = x^2 - x + 5, x > \frac{1}{2},$ and $g(x)$ is its inverse function,then $g'(7)$ equals

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