Figure 2 shows a sketch of the curve C with equation $y = f(x)$ where $$f(x) = 4(x^2 - 2)e^{-2x}$$ $x \\in \\mathbb{R}$ (a) Show that $f'(x) = 8(2 + x - x^2)e^{-2x}$ (b) Hence find, in simplest form, the exact coordinates of the stationary points of C - Edexcel - A-Level Maths Pure - Question 11 - 2020 - Paper 1

To find stationary points, we set $f'(x) = 0$:

1. Setting $8(2 + x - x^2)e^{-2x} = 0$ implies:

   • $2 + x - x^2 = 0$.

2. Rearranging leads to:

   • $x^2 - x - 2 = 0$.

3. Factoring the quadratic equation results in:

   • $(x - 2)(x + 1) = 0$.

4. Therefore, the solutions are $x = 2$ and $x = -1$.

5. To find the corresponding $y$-coordinates:

   • For $x = 2$:
     $f(2) = 4(2^2 - 2)e^{-4} = 4(2)e^{-4} = 8e^{-4}.$
   • For $x = -1$:
     $f(-1) = 4((-1)^2 - 2)e^{2} = 4(-1)e^{2} = -4e^{2}.$

6. The stationary points of C are thus $(-1, -4e^2)$ and $(2, 8e^{-4})$.

The function $g(x) = 2f(x)$ results in:

1. Since $f(x) = 4(x^2 - 2)e^{-2x}$ has a minimum at $(-1, -4e^2)$ and a maximum point, we will consider the behavior as $x \to \infty$ and $x \to -\infty$:

   • As $x \to \infty$, $f(x) \to 0$.
   • As $x \to -\infty$, $f(x) \to -\infty$.

2. Therefore, for $g$:

   • The minimum will be $2(-4e^2) = -8e^2$ and the maximum approaches $0$.

3. The range of $g$ is thus:

   $[-8e^2, 0).$

To find the range of $h(x) = 2f(x) - 3$:

1. The minimum value of $g(x)$ is $-8e^2$ so:

   • Substituting into $h$, we find:
   • Minimum: $-8e^2 - 3$.

2. As $g(x)$ approaches $0$, so:

   • Maximum: $0 - 3 = -3$.

3. Therefore, the overall range for $h$ is:

   $[-8e^2 - 3, -3).$