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Figure 2 shows a sketch of the curve C with equation $y = f(x)$ where $f(x) = 4(2x^2 - 2)e^{-2x}$ for $x \, \in \, \mathbb{R}$ - Edexcel - A-Level Maths Pure - Question 10 - 2020 - Paper 1

Question 10

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Figure 2 shows a sketch of the curve C with equation $y = f(x)$ where $f(x) = 4(2x^2 - 2)e^{-2x}$ for $x \, \in \, \mathbb{R}$. (a) Show that $f'(x) = 8(2 + x - x^... show full transcript

Worked Solution & Example Answer:Figure 2 shows a sketch of the curve C with equation $y = f(x)$ where $f(x) = 4(2x^2 - 2)e^{-2x}$ for $x \, \in \, \mathbb{R}$ - Edexcel - A-Level Maths Pure - Question 10 - 2020 - Paper 1

Step 1

Show that $f'(x) = 8(2 + x - x^2)e^{-2x}$

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Answer

To find the derivative of the function $f(x)$ , we utilize the product rule. We have:

$f(x) = 4(2x^2 - 2)e^{-2x}$

Letting $u = 4(2x^2 - 2)$ and $v = e^{-2x}$ , applying the product rule gives:

$f'(x) = u'v + uv'$

Calculating $u'$ , we find:

$u' = 8x$

Next, for $v'$ , we use the chain rule:

$v' = -2e^{-2x}$

Substituting these into our product rule:

$f'(x) = (8x)e^{-2x} + 4(2x^2 - 2)(-2e^{-2x})$

This simplifies to:

$f'(x) = 8xe^{-2x} - 8(2x^2 - 2)e^{-2x}$

Factoring out $8e^{-2x}$ leads to:

$f'(x) = 8e^{-2x}(x - (2x^2 - 2))$

Further simplification yields:

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

Step 2

Hence find, in simplest form, the exact coordinates of the stationary points of C.

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Answer

To find the stationary points, we solve for $f'(x) = 0$ :

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

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

$2 + x - x^2 = 0$

Rearranging gives:

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

Factoring the equation:

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

Thus, the solutions are:

$x = -2 \, \text{or} \, x = 1$

Substituting these values back into $f(x)$ to find the $y$ -coordinates:

For $x = -2$ : $f(-2) = 4(2(-2)^2 - 2)e^{4} = 4(8 - 2)e^{4} = 24e^{4}$
For $x = 1$ : $f(1) = 4(2(1^2) - 2)e^{-2} = 4(2 - 2)e^{-2} = 0$

Thus, the stationary points are $(-2, 24e^4)$ and $(1, 0)$ .

Step 3

Find (i) the range of g

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Answer

Given the function $g(x) = 2f(x)$ , we must determine its range:

Since $f(x) = 4(2x^2 - 2)e^{-2x}$ , we inspect the range of $f(x)$ :

The minimum value of $f(x)$ occurs at $x = 1$ , which gives $f(1) = 0$ .
As $x o - abla$ , $f(x) o abla$ .
As $x o abla$ , $f(x) o 0$ again.

Thus, the range of $f(x)$ is $[0, abla]$ . Consequently, the range of $g(x) = 2f(x)$ is $[0, abla]$ .

Step 4

Find (ii) the range of h

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Answer

To determine the range of $h(x) = 2f(x) - 3$ , we start with the range of $f(x)$ :

As established previously, the range is $[0, abla]$ . Therefore, applying the transformation for $h(x)$ leads to:

The minimum value will be $2(0) - 3 = -3$ .
The maximum does not change as $2f(x)$ expands without limit:

So the range of $h(x)$ is:

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Figure 2 shows a sketch of the curve C with equation $y = f(x)$ where $f(x) = 4(2x^2 - 2)e^{-2x}$ for $x \, \in \, \mathbb{R}$ - Edexcel - A-Level Maths Pure - Question 10 - 2020 - Paper 1

Worked Solution & Example Answer:Figure 2 shows a sketch of the curve C with equation $y = f(x)$ where $f(x) = 4(2x^2 - 2)e^{-2x}$ for $x \, \in \, \mathbb{R}$ - Edexcel - A-Level Maths Pure - Question 10 - 2020 - Paper 1

Show that $f'(x) = 8(2 + x - x^2)e^{-2x}$

Hence find, in simplest form, the exact coordinates of the stationary points of C.

Find (i) the range of g

Find (ii) the range of h

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