Key Concepts and Formulas

• Derivative as Slope of Tangent: The derivative of a function $f(x)$ at a point $x=c$, denoted by $f'(c)$, represents the slope of the tangent line to the curve $y=f(x)$ at the point $(c, f(c))$.
• Slope of a Line Segment: The slope of a line segment joining two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.
• Parallel Lines: Two lines are parallel if and only if they have the same slope.

Step-by-Step Solution

Step 1: Find the slope of the line segment joining the points (1, 0) and (e, e).

The slope of the line segment joining (1, 0) and (e, e) is given by: $m = \frac{e - 0}{e - 1} = \frac{e}{e - 1}$ This slope represents the slope that the tangent line must have.

Step 2: Find the derivative of the function $f(x) = x \log_e x$.

We need to find $f'(x)$. Using the product rule, we have: $f'(x) = \frac{d}{dx} (x \log_e x) = \frac{d}{dx}(x) \cdot \log_e x + x \cdot \frac{d}{dx}(\log_e x)$ $f'(x) = 1 \cdot \log_e x + x \cdot \frac{1}{x} = \log_e x + 1$

Step 3: Evaluate the derivative at $x = c$.

We have $f'(c) = \log_e c + 1$. This represents the slope of the tangent at the point $(c, f(c))$.

Step 4: Set the slope of the tangent equal to the slope of the line segment and solve for c.

Since the tangent at $(c, f(c))$ is parallel to the line segment joining (1, 0) and (e, e), their slopes must be equal. Therefore: $\log_e c + 1 = \frac{e}{e - 1}$ $\log_e c = \frac{e}{e - 1} - 1 = \frac{e - (e - 1)}{e - 1} = \frac{1}{e - 1}$ To solve for $c$, we exponentiate both sides with base $e$: $c = e^{\frac{1}{e - 1}}$

Step 5: Check the options

The options are: (A) ${{e - 1} \over e}$ (B) ${e^{\left( {{1 \over {1 - e}}} \right)}}$ (C) ${e^{\left( {{1 \over {e - 1}}} \right)}}$ (D) ${1 \over {e - 1}}$

Our derived answer is $c = e^{\frac{1}{e-1}}$ which matches option (C).

Common Mistakes & Tips

• Product Rule: Remember to apply the product rule correctly when differentiating $x \log_e x$.
• Parallel Lines: Understand that parallel lines have equal slopes.
• Logarithms and Exponentials: Be comfortable with the relationship between logarithms and exponentials to solve for $c$.
• Read the question carefully: Ensure that you answer the question asked.

Summary

The problem required us to find the value of $c$ such that the tangent to the curve $y = x \log_e x$ at $(c, f(c))$ is parallel to the line segment joining (1, 0) and (e, e). We found the slope of the line segment, then found the derivative of the function, evaluated it at $x=c$, and set the two slopes equal to each other. Solving for $c$ yielded $c = e^{\frac{1}{e - 1}}$, which corresponds to option (C). The correct answer provided was (A), which is incorrect.

Final Answer The final answer is \boxed{e^{\left( {{1 \over {e - 1}}} \right)}}, which corresponds to option (C).