Step-by-step Derivation

The problem asks for the location of the intersection point of a tangent line and a secant line to the curve $y = e^x$.

Let the function be $f(x) = e^x$. The three points mentioned in the problem are:

• The point of tangency: $P(c, e^c) = (c, f(c))$.
• The two points defining the secant line: $A(c-1, e^{c-1}) = (c-1, f(c-1))$ and $B(c+1, e^{c+1}) = (c+1, f(c+1))$.

We will determine the equations of the tangent and secant lines and find their intersection point.

Method 1: Using Properties of Convex Functions

1. Analyze the curve's convexity: The function is $f(x) = e^x$. The first derivative is $f'(x) = e^x$. The second derivative is $f''(x) = e^x$. Since $f''(x) = e^x > 0$ for all real $x$, the function $f(x)$ is strictly convex (or concave up).

2. Equation and Slope of the Tangent Line ($L_T$): The tangent is drawn at the point $(c, f(c))$. The slope of the tangent is $m_T = f'(c) = e^c$. The equation of the tangent line is $y_T(x) = f(c) + f'(c)(x-c) = e^c + e^c(x-c)$. At $x=c$, the y-value is $y_T(c) = e^c$.

3. Equation and Slope of the Secant Line ($L_S$): The secant line passes through $(c-1, f(c-1))$ and $(c+1, f(c+1))$. The slope of the secant is $m_S = \frac{f(c+1) - f(c-1)}{(c+1) - (c-1)} = \frac{e^{c+1} - e^{c-1}}{2}$. The equation of the secant line can be written as $y_S(x) = f(c-1) + m_S(x - (c-1))$.

4. Compare the Slopes: For a strictly convex function, the derivative $f'(x)$ is strictly increasing. By the Mean Value Theorem, the slope of the secant $m_S$ is equal to $f'(z)$ for some $z \in (c-1, c+1)$. A more direct comparison: $m_S = \frac{e^c(e - e^{-1})}{2} = e^c \left( \frac{e - 1/e}{2} \right) = e^c \sinh(1)$. Since $e \approx 2.718$, $e - 1/e > 2$, so $\sinh(1) = (e - 1/e)/2 > 1$. Therefore, $m_S = e^c \sinh(1) > e^c = m_T$. The secant line is steeper than the tangent line.

5. Compare the y-values at $x=c$: For the tangent line, $y_T(c) = e^c$. For the secant line, we find its y-value at the midpoint of the x-interval $[c-1, c+1]$: $y_S(c) = \frac{f(c-1) + f(c+1)}{2}$ (by linearity). $y_S(c) = \frac{e^{c-1} + e^{c+1}}{2} = e^c \frac{e^{-1} + e}{2} = e^c \cosh(1)$. For any strictly convex function, by Jensen's inequality, $\frac{f(a)+f(b)}{2} > f(\frac{a+b}{2})$. Here, $\frac{f(c-1)+f(c+1)}{2} > f(\frac{c-1+c+1}{2}) = f(c)$. So, $y_S(c) > y_T(c)$. At $x=c$, the secant line is above the tangent line.

6. Find the location of the intersection point $(x_{int}, y_{int})$: Let's set the equations of the lines equal: $y_T(c) + m_T(x_{int}-c) = y_S(c) + m_S(x_{int}-c)$ $(m_S - m_T)(x_{int}-c) = y_T(c) - y_S(c)$ We have established that:

   • $m_S - m_T > 0$ (a positive value).
   • $y_T(c) - y_S(c) < 0$ (a negative value). So, $(\text{positive}) \cdot (x_{int}-c) = (\text{negative})$. This implies that $(x_{int}-c)$ must be negative. $x_{int} - c < 0 \implies x_{int} < c$. The intersection point occurs to the left of $x=c$.

Method 2: Direct Algebraic Calculation

1. Tangent line equation: $y = e^c(x - c + 1)$.

2. Secant line equation: The slope is $m_S = \frac{e^{c+1} - e^{c-1}}{2}$. Using point $(c-1, e^{c-1})$: $y - e^{c-1} = \frac{e^{c+1} - e^{c-1}}{2} (x - (c-1))$.

3. Intersection: Equate the expressions for $y$: $e^c(x - c + 1) = e^{c-1} + \frac{e^{c+1} - e^{c-1}}{2} (x - c + 1)$. Let $X = x-c$. The equation becomes: $e^c(X+1) = e^{c-1} + \frac{e^{c+1} - e^{c-1}}{2} (X+1)$. $e^c X + e^c = e^{c-1} + \frac{e^{c+1}-e^{c-1}}{2}X + \frac{e^{c+1}-e^{c-1}}{2}$. $X \left( e^c - \frac{e^{c+1}-e^{c-1}}{2} \right) = e^{c-1} + \frac{e^{c+1}-e^{c-1}}{2} - e^c$. $X \left( \frac{2e^c - e^{c+1} + e^{c-1}}{2} \right) = \frac{2e^{c-1} + e^{c+1} - e^{c-1} - 2e^c}{2}$. $X \left( e^c(2 - e + e^{-1}) \right) = e^c(e^{-1} + e - 2)$. $X = \frac{e + e^{-1} - 2}{2 - e + e^{-1}}$. The numerator is $e + 1/e - 2 = (\sqrt{e} - 1/\sqrt{e})^2 > 0$ (since $e \neq 1$). The denominator is $2 - (e - e^{-1})$. Since $e \approx 2.718$, $e-e^{-1} \approx 2.35 > 2$. So the denominator is negative. Thus, $X = \frac{\text{positive}}{\text{negative}} < 0$. Since $X = x-c$, we have $x-c < 0$, which means $x < c$.

Both methods conclude that the intersection occurs to the left of $x=c$. This corresponds to option A.

Conclusion

The intersection point of the tangent line at $(c, e^c)$ and the secant line through $(c-1, e^{c-1})$ and $(c+1, e^{c+1})$ has an x-coordinate that is less than $c$. Therefore, the intersection is on the left of $x=c$.