Use the equation of the line tangent to $ f(x) = \frac{-2x}{x+1} $ at the point (0, 0) to approximate \(...

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Use the equation of the line tangent to $ f(x) = \frac{-2x}{x+1} $ at the point (0, 0) to approximate $ f(-0.1) $. Round answers expressed as decimals to the nearest thousandth. $ f(-0.1) \approx $

SnapAndSolve AI · Accepted Answer

To find the equation of the tangent line, we first need the derivative of $ f(x) = \frac{-2x}{x+1} $.

The derivative $ f'(x) $ is found using the quotient rule:

\[ f'(x) = \frac{(x+1)(-2) - (-2x)(1)}{(x+1)^2} = \frac{-2x - 2 + 2x}{(x+1)^2} = \frac{-2}{(x+1)^2} \]

At the point (0, 0), $ f'(0) = \frac{-2}{(0+1)^2} = -2 $.

The equation of the tangent line at (0, 0) is:

\[ y - 0 = -2(x - 0) \]

\[ y = -2x \]

To approximate $ f(-0.1) $, substitute $ x = -0.1 $ into the tangent line equation:

\[ y = -2(-0.1) = 0.2 \]

Therefore, $ f(-0.1) \approx 0.2 $.

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Use the equation of the line tangent to \( f(x) = \frac{-2x}{x+1} \) at the point (0, 0) to approximate \(...

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