The tangent to the curve y = e^x at the point | vedclass

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Question

(A) Given the curve $y = e^x$,the slope of the tangent at $(c, e^c)$ is given by $\frac{dy}{dx} = e^x$. At $x = c$,the slope $m_1 = e^c$. The equation

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less than $c$.

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(A) Given the curve $y = e^x$,the slope of the tangent at $(c, e^c)$ is given by $\frac{dy}{dx} = e^x$. At $x = c$,the slope $m_1 = e^c$. The equation of the tangent at $(c, e^c)$ is $y - e^c = e^c(x - c)$,which simplifies to $y = e^c(x - c + 1)$. Now,the slope of the line joining $(c - 1, e^{c-1})$ and $(c + 1, e^{c+1})$ is $m_2 = \frac{e^{c+1} - e^{c-1}}{(c+1) - (c-1)} = \frac{e^{c-1}(e^2 - 1)}{2}$. The equation of this secant line is $y - e^{c-1} = m_2(x - (c - 1))$. To find the intersection,we set the $y$-values equal: $e^c(x - c + 1) = e^{c-1}(x - c + 1) + e^{c-1}$. Solving for $x$,we find that the $x$-coordinate of the intersection is $c - 1 + \frac{2}{e^2 - 1} \cdot e$. Since $e \approx 2.718$,$e^2 - 1 \approx 6.389$. Thus,$x = c - 1 + \frac{2e}{e^2 - 1} Therefore,the intersection point has an $x$-coordinate less than $c$.

The tangent to the curve $y = e^x$ at the point $(c, e^c)$ intersects the line joining the points $(c - 1, e^{c-1})$ and $(c + 1, e^{c+1})$ at a point whose $x$-coordinate is:

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