Consider the graph of $y = e^x - x$ and a horizontal line $y = k$. How is the number of real solutions related to how many times the line $y = k$ intersects the graph of $y = e^x - x$?

Since $e^x - x$ becomes very large for both very negative and very positive $x$, the graph has a lowest point (a minimum). For which value of $k$ would a horizontal line just touch the graph at that minimum instead of crossing it twice or missing it entirely?

Let $f(x)=e^x - x$. If you know derivatives, set $f'(x)=0$ to find the minimum. If you prefer graphing, plot $y=e^x - x$ on a calculator or Desmos and identify its lowest $y$-value. That value will tell you which $k$ gives exactly one solution.

Enter y1 = e^x to graph the exponential curve. Then, for each answer choice for $k$, enter a separate line: y2 = x - 1, y3 = x, y4 = x + 1, and y5 = x + 2.

For each line, use Desmos’s intersection tool (tap on where the graphs meet or use the point-of-intersection feature) to see how many intersection points it has with the curve $y=e^x$. The correct $k$ is the one whose line intersects the curve at exactly one point.

Start by moving $x$ to the left side:

Now think of $y = e^x - x$ as a graph and $y = k$ as a horizontal line. Each solution to the equation corresponds to an $x$-value where the graph of $y = e^x - x$ intersects the horizontal line $y = k$.

As $x \to -\infty$, $e^x \to 0$ and $-x \to +\infty$, so $e^x - x$ becomes very large and positive. As $x \to +\infty$, $e^x$ also becomes very large, so $e^x - x$ again becomes very large and positive.

This means the graph of $y = e^x - x$ goes up to $+\infty$ on both the far left and far right, so it must dip down to a lowest point (a minimum) in between. A horizontal line $y = k$ will:

• not intersect the graph if $k$ is less than this minimum,
• intersect it exactly once if $k$ equals this minimum,
• intersect it twice if $k$ is greater than this minimum.

So we need the minimum value of $e^x - x$ to know which $k$ gives exactly one solution.

Let $f(x) = e^x - x$. To find its minimum (if you know derivatives), compute:

Set this equal to $0$ to find critical points:

This happens at $x = 0$. Since $f(x)$ goes to $+\infty$ at both ends and $f(x)$ is smooth, this critical point at $x=0$ is the minimum.

Now evaluate $f$ at $x=0$:

So the minimum value of $e^x - x$ is $1$.

From $e^x - x = k$, the number $k$ is exactly the $y$-value where the horizontal line $y = k$ intersects the graph of $y = e^x - x$.

We found that the smallest possible value of $e^x - x$ is $1$. Therefore:

• if $k < 1$, there is no solution,
• if $k = 1$, the line just touches the graph at that minimum point, giving exactly one real solution,
• if $k > 1$, the line cuts the graph in two points, giving two real solutions.

Since the question asks for the value of $k$ that gives exactly one real solution, the correct choice is $k = 1$ (answer choice C).