The coordinates of the point P on the graph o | vedclass

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(C) Let the point $P$ be $(x_0, y_0)$ where $x_0 > 0$. Then $y_0 = e^{-x_0}$.The slope of the tangent at $P$ is $m = \frac{dy}{dx} = -e^{-x_0}$.The eq

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Both $(A)$ and $(B)$

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(C) Let the point $P$ be $(x_0, y_0)$ where $x_0 > 0$. Then $y_0 = e^{-x_0}$.The slope of the tangent at $P$ is $m = \frac{dy}{dx} = -e^{-x_0}$.The equation of the tangent is $y - e^{-x_0} = -e^{-x_0}(x - x_0)$.For $x$-intercept,set $y = 0$: $-e^{-x_0} = -e^{-x_0}(x - x_0) \implies x = x_0 + 1$.For $y$-intercept,set $x = 0$: $y = e^{-x_0} + x_0 e^{-x_0} = e^{-x_0}(1 + x_0)$.The area of the triangle formed by the tangent and the axes is $A = \frac{1}{2} |x_{int}| |y_{int}| = \frac{1}{2} (x_0 + 1) (e^{-x_0}(1 + x_0)) = \frac{1}{2} e^{-x_0} (x_0 + 1)^2$.To maximize $A$,find $\frac{dA}{dx_0} = \frac{1}{2} [ -e^{-x_0}(x_0 + 1)^2 + e^{-x_0} \cdot 2(x_0 + 1) ] = \frac{1}{2} e^{-x_0} (x_0 + 1) [ -x_0 - 1 + 2 ] = \frac{1}{2} e^{-x_0} (x_0 + 1) (1 - x_0)$.Setting $\frac{dA}{dx_0} = 0$ gives $x_0 = 1$.Thus,the point is $(1, 1/e)$.Since $y = e^{-|x|}$ is an even function,the graph is symmetric about the $y$-axis,so the point $(-1, 1/e)$ also yields the same maximum area.Therefore,both $(A)$ and $(B)$ are correct.

The coordinates of the point $P$ on the graph of the function $y = e^{-|x|}$ where the portion of the tangent intercepted between the coordinate axes has the greatest area,is

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