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

on the left of x = c

The equation of the tangent to the curve $y=e^x$ at $\left(c, e^c\right)$ is

$y-e^c=e^c(x-c)$              ..........(i)

The equation of the line joining $\left(c-1, e^{c-1}\right)$ and $\left(c+1, e^{c+1}\right)$ is

$y-e^{c-1}=\frac{e^c\left(e-e^{-1}\right)}{2}(x-c+1)$              ..........(ii)

On subtracting (i) from (ii), we get

$e^c-e^{c-1}=\frac{e^c\left(e-e^{-1}\right)}{2}(x-c+1)-e^c(x-c)$

$\Rightarrow e^c-e^{c-1}=(x-c) e^c\left(\frac{e-e^{-1}}{2}-1\right)+e^c\left(\frac{e-e^{-1}}{2}\right)$

$\Rightarrow x-c=\frac{(e-1)^2}{2-(e-1)^2}<0 \Rightarrow x<c$

Hence, (i) and (ii) intersect at a point on the left of x = c.

ALITER   

Clearly, point $P\left(c, e^c\right)$ lies in between the points $Q\left(c-1, e^{c-1}\right)$ and $R\left(c+1, e^{c+1}\right)$ on the curve $y=e^x$ as shown in Figure. Also, $y=e^x$ is strictly increasing. So, the tangent at P.