Figure 2 shows a sketch of part of the curve with equation $$g(x) = x^2(1 - x)e^{-2x}, \, x > 0$$ (a) Show that $g'(x) = f(x)e^{-2x}$, where $f(x)$ is a cubic function to be found - Edexcel - A-Level Maths Pure - Question 8 - 2015 - Paper 3

To find the range of the function $g(x) = x^2(1 - x)e^{-2x}$, we analyze the behavior of $g(x)$ as follows:

1. Find critical points by setting the derivative $g'(x)$ to zero:

   $g'(x) = e^{-2x}(2x^3 - 3x^2) = 0$ This implies: $2x^3 - 3x^2 = 0$ Factor to get: $x^2(2x - 3) = 0$ This yields critical points at $x = 0$ and x = rac{3}{2}.

2. Evaluate $g(x)$ at these points:

   • At $x = 0$: $g(0) = 0^2(1 - 0)e^{-0} = 0$
   • At x = rac{3}{2}: gigg(rac{3}{2}igg) = igg(rac{3}{2}igg)^2igg(1 - rac{3}{2}igg)e^{-3} = rac{9}{4}igg(-rac{1}{2}igg)e^{-3} = -rac{9}{8}e^{-3}

   Since $g(x) o 0$ as $x o ext{large}$, we will next find if there is a maximum value.

3. Consider the behaviour as $x o ext{large}$:

   • As $x$ increases, $g(x)$ approaches $0$ from the positive side.
   • Hence, the function achieves a maximum somewhere between the critical points.

Thus, the range of $g(x)$ is [0, -rac{9}{8}e^{-3}] for $x > 0$.

The function $g(x)$ is not one-to-one (injective) since it has multiple $x$ values producing the same $y$ output.

1. Analyzing the graph: From the graph in Figure 2, it is clear that $g(x)$ is a many-to-one function due to the presence of turning points.

2. Conclusion: Hence, $g^{-1}(y)$ does not exist because it doesn't satisfy the criteria for invertibility where every output $y$ corresponds to exactly one input $x$.