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

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(A) The slope of the line joining the points $(c-1, e^{c-1})$ and $(c+1, e^{c+1})$ is given by $m = \frac{e^{c+1}-e^{c-1}}{(c+1)-(c-1)} = \frac{e^{c+1

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on the left of $x=c$

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(A) The slope of the line joining the points $(c-1, e^{c-1})$ and $(c+1, e^{c+1})$ is given by $m = \frac{e^{c+1}-e^{c-1}}{(c+1)-(c-1)} = \frac{e^{c+1}-e^{c-1}}{2}$.Since $e^x$ is a strictly convex function,the slope of the secant line between any two points is greater than the slope of the tangent at any point between them. Specifically,$\frac{e^{c+1}-e^{c-1}}{2} > e^c$ because $\frac{e-e^{-1}}{2} > 1$ (since $e \approx 2.718$,$\frac{2.718-0.368}{2} = 1.175 > 1$).Because the slope of the secant line is greater than the slope of the tangent at $(c, e^c)$,and the function is convex,the tangent line must lie below the secant line for $x > c$ and above it for $x Alternatively,the equation of the tangent at $(c, e^c)$ is $y-e^c = e^c(x-c)$,or $y = e^c(x-c+1)$.The equation of the secant line is $y-e^{c-1} = \frac{e^{c+1}-e^{c-1}}{2}(x-(c-1))$.Solving these simultaneously for $x$ yields $x < c$.

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

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