Question: User: Question 5
(a) On a single set of axes, sketch the curves of $y = e^x - 1$ and $y = 6e^{-x}$.
(b) Show that the coordinates of the point of intersection of these curves is the
equation $e^{2x} - e^x - 6 = 0$.
(c) Hence find the exact coordinates of the point of intersection of these curves.
Graph shapes: two exponential curves on the same axes; $y = e^x - 1$ is an increasing exponential shifted down by 1, crossing the y-axis at $(0, -1)$ and the x-axis at $(0, 0)$, while $y = 6e^{-x}$ is a decreasing exponential crossing the y-axis at $(0, 6)$; they intersect in the center-right region. position_hint = center
1. **State the problem:** We are given two curves $y = e^x - 1$ and $y = 6e^{-x}$ and asked to (a) sketch them, (b) show the intersection satisfies $e^{2x} - e^x - 6 = 0$, and (c) find the exact intersection coordinates.
2. **Sketching the curves:**
- The curve $y = e^x - 1$ is an exponential growth curve shifted down by 1.
- It crosses the y-axis at $y = e^0 - 1 = 1 - 1 = 0$ (point $(0,0)$).
- The curve $y = 6e^{-x}$ is a decreasing exponential curve.
- It crosses the y-axis at $y = 6e^0 = 6$ (point $(0,6)$).
3. **Find the intersection:** At the point(s) of intersection, the $y$-values are equal:
$$e^x - 1 = 6e^{-x}$$
4. **Rewrite the equation:** Multiply both sides by $e^x$ to clear the negative exponent:
$$e^x(e^x - 1) = 6e^x e^{-x}$$
$$e^{2x} - e^x = 6$$
5. **Rearrange to standard form:**
$$e^{2x} - e^x - 6 = 0$$
This confirms the equation for the intersection point.
6. **Solve the quadratic in $e^x$:** Let $u = e^x$, then
$$u^2 - u - 6 = 0$$
7. **Factor or use quadratic formula:**
$$u = \frac{1 \pm \sqrt{1 + 24}}{2} = \frac{1 \pm 5}{2}$$
8. **Find roots:**
- $u = \frac{1 + 5}{2} = 3$
- $u = \frac{1 - 5}{2} = -2$ (discard since $e^x > 0$ always)
9. **Find $x$:**
$$e^x = 3 \implies x = \ln 3$$
10. **Find $y$ coordinate:** Substitute back into one of the original equations, e.g., $y = e^x - 1$:
$$y = 3 - 1 = 2$$
**Final answer:** The curves intersect at exactly one point with coordinates
$$\boxed{\left(\ln 3, 2\right)}$$