If f(x) = frac{x^2-x}{x^2+2x},then the value | vedclass

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(B) Given $f(x) = \frac{x^2-x}{x^2+2x}$. For $x 
eq 0$,we can simplify this as $f(x) = \frac{x-1}{x+2}$.Let $y = \frac{x-1}{x+2}$.To find $f^{-1}(x)$

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Given $f(x) = \frac{x^2-x}{x^2+2x}$. For $x 
eq 0$,we can simplify this as $f(x) = \frac{x-1}{x+2}$.Let $y = \frac{x-1}{x+2}$.To find $f^{-1}(x)$,we solve for $x$ in terms of $y$:$y(x+2) = x-1$$yx + 2y = x - 1$$yx - x = -1 - 2y$$x(y-1) = -(1+2y)$$x = \frac{2y+1}{1-y}$.Thus,$f^{-1}(x) = \frac{2x+1}{1-x}$.Now,we differentiate $f^{-1}(x)$ with respect to $x$ using the quotient rule:$\frac{d}{dx}(f^{-1}(x)) = \frac{d}{dx}\left(\frac{2x+1}{1-x}ight) = \frac{(1-x)(2) - (2x+1)(-1)}{(1-x)^2}$$= \frac{2 - 2x + 2x + 1}{(1-x)^2} = \frac{3}{(1-x)^2}$.Evaluating at $x = 2$:$\frac{d}{dx}(f^{-1}(x))|_{x=2} = \frac{3}{(1-2)^2} = \frac{3}{(-1)^2} = 3$.

If $f(x) = \frac{x^2-x}{x^2+2x}$,then the value of $\frac{d}{dx}(f^{-1}(x))$ at $x = 2$ is:

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