Figure 2 shows a sketch of part of the curve with equation g(r) = r^2(1 - r)e^{-2r}, \, r > 0 (a) Show that g'(r) = f(c) e^{-2r}, where f(c) is a cubic function to be found - Edexcel - A-Level Maths Pure - Question 9 - 2015 - Paper 3

To find the range of g, we analyze the behavior of the function: $g(r) = r^2(1 - r)e^{-2r}$ for $r > 0$.

1. As $r ightarrow 0$, $g(0) = 0$.
2. As $r ightarrow ext{large values}$, since $e^{-2r}$ approaches $0$, $g(r)$ also approaches $0$.
3. To find critical points, we can set $g'(r) = 0$, which simplifies to finding the roots of: $2r - 5r^2 + 2r^3 = 0$ Factoring gives: $r(2 - 5r + 2r^2) = 0$ Solving the quadratic part: $2r^2 - 5r + 2 = 0$ Using the quadratic formula $r = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4}$ gives:
4. $r = 2$, which is in the relevant domain, and
5. $r = 0.5$, also valid.

By evaluating $g(2)$ and $g(0.5)$, we find:

• $g(2) = 0$
• $g(0.5) = g(0.5) = \frac{0.5^2(1 - 0.5)}{e^1} = \frac{0.25 \times 0.5}{e} = \frac{0.125}{e}$, indicating that the maximum value occurs at $r = 0.5$.

Thus, the range of g is: $0 < g(r) < \frac{0.125}{e}$.

The function g^{-1}(r) does not exist because g(r) is a many-to-one function. This is due to the fact that g(r) reaches the same values for different input values (as seen in the critical points and the behavior of the function as it approaches values). Hence, the inverse cannot be uniquely defined.