If the line y=-4x+b is tangent to the curve y | vedclass

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(A) The slope of the line $y=-4x+b$ is $m=-4$. The slope of the tangent to the curve $y=\frac{1}{x}$ is given by the derivative $\frac{dy}{dx} = -\fra

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$\pm 4$

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(A) The slope of the line $y=-4x+b$ is $m=-4$. The slope of the tangent to the curve $y=\frac{1}{x}$ is given by the derivative $\frac{dy}{dx} = -\frac{1}{x^2}$. Since the line is tangent to the curve,their slopes must be equal at the point of tangency: $-\frac{1}{x^2} = -4 \Rightarrow x^2 = \frac{1}{4} \Rightarrow x = \pm \frac{1}{2}$. For $x = \frac{1}{2}$,the $y$-coordinate on the curve is $y = \frac{1}{1/2} = 2$. For $x = -\frac{1}{2}$,the $y$-coordinate on the curve is $y = \frac{1}{-1/2} = -2$. Substituting these points $(x, y)$ into the line equation $y = -4x + b$: Case $1$: $2 = -4(\frac{1}{2}) + b \Rightarrow 2 = -2 + b \Rightarrow b = 4$. Case $2$: $-2 = -4(-\frac{1}{2}) + b \Rightarrow -2 = 2 + b \Rightarrow b = -4$. Thus,$b = \pm 4$.

If the line $y=-4x+b$ is tangent to the curve $y=\frac{1}{x}$,then $b$ equals

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