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 10 - 2020 - Paper 1

Stationary points occur where $f'(x) = 0$. Setting the derivative equal to zero gives us:

$8(2 + x - x^2)e^{-2x} = 0.$ Since $e^{-2x}$ is never zero, we have:

$8(2 + x - x^2) = 0,$ which simplifies to:

$2 + x - x^2 = 0.$ Rearranging gives:

$x^2 - x - 2 = 0.$ Factoring yields: $(x - 2)(x + 1) = 0.$ Thus, the solutions are:

$x = 2 \\quad ext{and} \\quad x = -1.$ Now we find the corresponding $y$ values:

• For $x = 2$: $f(2) = 4(2^2 - 2)e^{-4} = 4(4 - 2)e^{-4} = 8e^{-4}.$
• For $x = -1$: $f(-1) = 4((-1)^2 - 2)e^{2} = 4(1 - 2)e^{2} = -4e^{2}.$ Thus, the stationary points are $(-1, -4e^2)$ and $(2, 8e^{-4})$.

To find the range of $g(x) = 2f(x)$, we first examine $f(x) = 4(x^2 - 2)e^{-2x}$. Given the stationary points from part (b), we need to analyze $f(x)$ further:

1. For $x < -1$, $f(x)$ approaches $0$ from below, and as $x$ approaches $-1$, $f(-1)$ gives $-4e^2$.
2. The function increases to a maximum at $x = 2$, where $f(2) = 8e^{-4}$.
3. As $x o ext{infinity}$, $f(x) o 0$.

Thus, the range of $f(x)$ is approximately $[-4e^{2}, 8e^{-4}]$. Therefore, the range of $g(x)$ is: $g(x) \\in [2(-4e^{2}), 2(8e^{-4})] = [-8e^{2}, 16e^{-4}].$

The function $h(x) = 2f(x) - 3$. Given the range of $f(x)$ as $[-4e^{2}, 8e^{-4}]$, we need to shift this range:

1. The lower limit becomes: $2(-4e^{2}) - 3 = -8e^{2} - 3.$
2. The upper limit becomes: $2(8e^{-4}) - 3 = 16e^{-4} - 3.$

Thus, the range of $h(x)$ is: $h(x) \\in [-8e^{2} - 3, 16e^{-4} - 3].$