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 $f'(x) = 8(2 + x - x^2)e^{-2x}$ - Edexcel - A-Level Maths Pure - Question 9 - 2020 - Paper 1

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

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

Since $e^{-2x}$ is never zero, we focus on:

$2 + x - x^2 = 0$

Rearranging the equation leads to:

$x^2 - x - 2 = 0$

Factoring gives:

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

This results in:

• $x = 2$
• $x = -1$

To find the corresponding $y$ coordinates:

• 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:

• $(-1, -4e^2)$
• $(2, 8e^{-4})$

Given $g(x) = 2f(x)$, we first find the range of $f(x)$:

We know that the stationary points are:

• $(-1, -4e^2)$ with $f(-1) = -4e^2$
• $(2, 8e^{-4})$ with $f(2) = 8e^{-4}$

From the behavior of the function as $x \to \pm \infty$, we observe:

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

Thus, the range of $f(x)$ is:

• $[-4e^2, 8e^{-4}]$

Hence, the range of $g(x)$ is:

• $[2(-4e^2), 2(8e^{-4})] = [-8e^2, 16e^{-4}]$.

Given $h(x) = 2f(x) - 3$ with $y > 0$, the range can be found from the range of $f(x)$:

We already know:

• The range of $f(x)$ is $[-4e^2, 8e^{-4}]$.

Subtracting 3 from each part of the range of $f(x)$ yields:

• Lower bound: $-4e^2 - 3$
• Upper bound: $8e^{-4} - 3$

Thus, the range of $h(x)$ is:

• $[-4e^2 - 3, 8e^{-4} - 3]$.
• Considering $y > 0$, we determine valid values for this result.