Intersection of Tangent to Exponential Curve with Secant Line

To find the point of intersection between the tangent line and the secant line, we first determine the equations of both lines.

1. Equation of the Tangent Line:
   The given curve is $y = e^x$. The derivative with respect to $x$ is:
   $\frac{dy}{dx} = e^x$
   At the point $(c, e^c)$, the slope of the tangent is:
   $m_1 = \left.\frac{dy}{dx}\right|_{x=c} = e^c$
   Using the point-slope form, the equation of the tangent line is:
   $y - e^c = e^c(x - c) \implies y = e^c(x - c + 1)$

2. Equation of the Secant Line:
   The line joins the points $(c-1, e^{c-1})$ and $(c+1, e^{c+1})$. The slope of this secant line is:
   $m_2 = \frac{e^{c+1} - e^{c-1}}{(c+1) - (c-1)} = \frac{e^c(e - e^{-1})}{2}$
   Using the point $(c-1, e^{c-1})$, the equation of the secant line is:
   $y - e^{c-1} = m_2(x - (c-1))$
   $y = \frac{e^c(e - e^{-1})}{2}(x - c + 1) + e^{c-1}$

3. Finding the Point of Intersection:
   Equating the two expressions for $y$:
   $e^c(x - c + 1) = \frac{e^c(e - e^{-1})}{2}(x - c + 1) + e^{c-1}$
   Dividing the entire equation by $e^c$:
   $(x - c + 1) = \left(\frac{e - e^{-1}}{2}\right)(x - c + 1) + e^{-1}$
   Let $u = x - c$. Then:
   $(u + 1)\left(1 - \frac{e - e^{-1}}{2}\right) = e^{-1}$
   $(u + 1)\left(\frac{2 - e + e^{-1}}{2}\right) = \frac{1}{e}$
   Solving for $u + 1$:
   $u + 1 = \frac{2}{e(2 - e + e^{-1})} = \frac{2}{2e - e^2 + 1}$

Now, let us evaluate the denominator:
$2e - e^2 + 1 = 1 - (e - 1)^2 + 1 = 2 - (e - 1)^2$
Since $e \approx 2.718$, we have:
$(e - 1)^2 \approx (1.718)^2 \approx 2.952 > 2$
Therefore:
$2e - e^2 + 1 < 0$
Since the numerator is positive ($2 > 0$) and the denominator is negative, we have:
$u + 1 < 0 \implies u < -1$
Substituting back $u = x - c$:
$x - c < -1 \implies x < c - 1 < c$

Hence, the $x$-coordinate of the point of intersection satisfies $x < c$, which means the intersection lies on the left of $x = c$.

Therefore, the correct option is (A).