Figure 2 shows a sketch of the curve C with equation $y = f(x)$ where $$f(x) = 4igg(x^2 - 2igg)e^{-2x}$$, $x \in \mathbb{R}$ - Edexcel - A-Level Maths Pure - Question 9 - 2020 - Paper 1

The stationary points occur where $f^{\prime}(x) = 0$:

$8(2 + x - x^2)e^{-2x} = 0$

Since $e^{-2x} > 0$ for all $x$, we focus on:

$2 + x - x^2 = 0 \\ x^2 - x - 2 = 0$

Factoring gives:

$(x - 2)(x + 1) = 0$

Thus, the roots are:

• $x = 2$
• $x = -1$

Calculating the $y$-coordinates:

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

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

Thus, the exact coordinates of the stationary points are:

• $(-1, -4e^2)$
• $(2, 8e^{-4})$

First, we know:

$g(x) = 2f(x)$

Substituting $f(x)$ we have:

$g(x) = 2 imes 4(x^2 - 2)e^{-2x} = 8(x^2 - 2)e^{-2x}$

As $x$ goes to positive or negative infinity, the term $e^{-2x}$ approaches $0$ making $g(x)$ approach $0$. Hence, we need to evaluate the minimum and maximum values of $g(x)$ at stationary points:

At $x = -1$ and $x = 2$, we already computed:

• $g(-1) = 2(-4e^2) = -8e^2$
• $g(2) = 2(8e^{-4}) = 16e^{-4}$

Thus, the range of $g$ is:

$[-8e^2, 16e^{-4}]$

Using the definition of $h$:

$h(x) = 2f(x) - 3$

Thus:

$h(x) = 8(x^2 - 2)e^{-2x} - 3$

To find the range of $h(x)$, we look at the minimum and maximum of $g(x)$ and adjust for the constant:

The minimum of $g$ is $-8e^2$, giving:

$h(-1) = -8e^2 - 3$

The maximum value of $g$ is $16e^{-4}$, thus:

$h(2) = 16e^{-4} - 3$

The range of $h$ then becomes:

$[-8e^2 - 3, 16e^{-4} - 3]$