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

Slope of the line joining the points \(\left(c-1, e^{c-1}\right)\) and \(\left(c+1, e^{c+1}\right)\) is equal to \(\frac{e^{c+1}-e^{c-1}}{2}>e^c\)


\[
\text { Tangent to the curve } y=e^x \text { will intersect the given line to the left of the line } x=c \text {. }
\]
ALITER
The equation of the tangent to the curve \(y=e^x\) at \(\left(c, e^c\right)\) is
\[
y-e^c=e^c(x-c)
\]
Equation of the line joining the given points is
\[
y-e^{c-1}=\frac{e^c\left(e-e^{-1}\right)}{2}[x-(c-1)]
\]
Eliminating \(y\) from Eqs. (i) and (ii), we get
\[
\begin{aligned}
{[x-(c-1)]\left[2-\left(e-e^{-1}\right)\right] } & =2 e^{-1} \\
x-c & =\frac{e+e^{-1}-2}{2-\left(e-e^{-1}\right)} < 0 \Rightarrow x < c .
\end{aligned}
\]