Animated Solution for Mathematics - Differentiation: Comprehension Passage

We are tasked with finding the roots of the function $f(x)=k{e}^x-x$. Because the presence of $e^x$ makes this a transcendental equation, we cannot isolate $x$ using basic algebraic manipulation.

Instead, we shift our perspective to the geometric interpretation. We rewrite the equation $k{e}^x-x=0$ as:

The roots of the original equation correspond to the $x$-coordinates where the line $y_1=x$ intersects the exponential curve $y_2=k{e}^x$.

Consider the case where $k\le 0$. If $k=0$, the curve $y=k{e}^x$ collapses into the $x$-axis ($y=0$), which intersects the line $y=x$ at exactly one point: the origin $(0,0)$.

If $k<0$, the curve $y=k{e}^x$ lies entirely below the $x$-axis. As $x\to -\infty$, $k{e}^x\to 0$, and as $x\to \infty$, $k{e}^x\to -\infty$.

Since the line $y=x$ also extends into the third quadrant, the two graphs are guaranteed to intersect. Thus, for all $k\le 0$, there is exactly one root.

When $k>0$, the exponential curve is strictly positive. To find the transition point between zero, one, and two roots, we identify the condition of tangency.

At the point of tangency $(x_0,y_0)$, two conditions must be satisfied:

1. The functions must intersect: $k{e}^{x_0}=x_0$ 2. Their slopes must be equal: $\frac{d}{dx}(k{e}^x)=\frac{d}{dx}(x)⟹k{e}^{x_0}=1$

Substituting the second condition into the first, we find $x_0=1$. Substituting $x_0=1$ back into the slope equation yields:

This value, $k=\frac{1}{e}$, represents the critical threshold for the existence of roots.

When $k$ lies in the interval $(0,\frac{1}{e})$, the exponential curve is sufficiently "low" to cross the line $y=x$ at two distinct points.

If $k>\frac{1}{e}$, the curve $y=k{e}^x$ lifts entirely above the line $y=x$, resulting in zero roots. If $k=\frac{1}{e}$, the graphs are tangent, resulting in exactly one root.

Therefore, the equation $k{e}^x-x=0$ yields two distinct roots when $k\in (0,\frac{1}{e})$.