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

$\begin{array}{l}\text{ Equation of straight line joining }A\left(c+1,e^{c+1}\right)\\ \text{ and }B\left(c-1,e^{c-1}\right)\text{ is }\\ y-e^{c+1}=\frac{e^{c+1}-e^{c-1}}{2}(x-c-1)-----\left(1\right)\end{array}$

$\text{ Equation of tangent at }\left(c,e^c\right)\text{ is }$$y-e^c=e^c(x-c)----\left(2\right)$

Subtracting (1) from (2),  we get

$\begin{array}{l}{e}^c(e-1)=e^c\left[(x-c)-\frac{1}{2}\left(e-e^{-1}\right)(x-c)+\frac{1}{2}\left(e-e^{-1}\right)\right]\\ \Rightarrow \frac{1}{2}\left(e+e^{-1}\right)-1=(x-c)\left[1-\frac{1}{2}\left(e-e^{-1}\right)\right]\\ \Rightarrow x-c=\frac{e+e^{-1}-2}{2-e+e^{-1}}<0\\ \left[\therefore e+e^{-1}>2\text{ and }2+e^{-1}-e<0\right]\end{array}$

$\begin{array}{l}\Rightarrow x<c\\ \text{ Thus the two lines meet to the left of }\mathrm{x}=\mathrm{c}\text{. }\end{array}$