The curve with equation $y = f(x) = 3x e^x - 1$ has a turning point $P$. (a) Find the exact coordinates of $P$. (b) The equation $f(x) = 0$ has a root between $x = 0.25$ and $x = 0.3$. (c) Use the iterative formula $$x_{n+1} = \frac{1}{3} e^{-x_n}$$ with $x_0 = 0.25$ to find, to 4 decimal places, the values of $x_1$, $x_2$, and $x_3$. By choosing a suitable interval, show that a root of $f(x) = 0$ is $x = 0.2576$ correct to 4 decimal places.

A-Level Edexcel Maths Pure Trigonometric Equations: The curve with equation $y = f(x

The curve with equation $y = f(x) = 3x e^x - 1$ has a turning point $P$ - Edexcel - A-Level Maths Pure - Question 8 - 2009 - Paper 2

Question 8

The curve with equation $y = f(x) = 3x e^x - 1$ has a turning point $P$. (a) Find the exact coordinates of $P$. (b) The equation $f(x) = 0$ has a root between $x =... show full transcript

Worked Solution & Example Answer:The curve with equation $y = f(x) = 3x e^x - 1$ has a turning point $P$ - Edexcel - A-Level Maths Pure - Question 8 - 2009 - Paper 2

Step 1

Find the exact coordinates of P.

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Answer

To find the turning point $P$ , we need to calculate the first derivative of the function:

$f'(x) = 3 e^x + 3x e^x$

Setting $f'(x) = 0$ :

$3 e^x (1 + x) = 0$

As $e^x$ is never zero, we focus on the equation $1 + x = 0$ . Thus, we find:

$x = -1$

Next, we calculate the corresponding $y$ coordinate:

$f(-1) = 3(-1)e^{-1} - 1 = -\frac{3}{e} - 1$

Hence, the coordinates of $P$ are:

$\left(-1, -\frac{3}{e} - 1\right)$

Step 2

Use the iterative formula with x_0 = 0.25 to find, to 4 decimal places, the values of x_1, x_2, and x_3.

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Answer

Using the iterative formula:

$x_{n+1} = \frac{1}{3} e^{-x_n}$

For $x_0 = 0.25$ :

$x_1 = \frac{1}{3} e^{-0.25} \approx 0.2596$
For $x_1 = 0.2596$ :

$x_2 = \frac{1}{3} e^{-0.2596} \approx 0.2571$
For $x_2 = 0.2571$ :

$x_3 = \frac{1}{3} e^{-0.2571} \approx 0.2578$

Thus, to 4 decimal places:

$x_1 \approx 0.2596$
$x_2 \approx 0.2571$
$x_3 \approx 0.2578$

Step 3

By choosing a suitable interval, show that a root of f(x) = 0 is x = 0.2576 correct to 4 decimal places.

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Answer

To show that a root exists between $0.2575$ and $0.25765$ , we evaluate the function:

For $f(0.2575)$ :

$f(0.2575) = 3(0.2575)e^{0.2575} - 1 \approx 0.000379...$
For $f(0.25765)$ :

$f(0.25765) = 3(0.25765)e^{0.25765} - 1 \approx -0.000109...$

The change of sign indicates a root exists in $(0.25755, 0.25765)$ , confirming: $x \approx 0.2576$ is accurate to 4 decimal places.

The curve with equation $y = f(x) = 3x e^x - 1$ has a turning point $P$ - Edexcel - A-Level Maths Pure - Question 8 - 2009 - Paper 2

Worked Solution & Example Answer:The curve with equation $y = f(x) = 3x e^x - 1$ has a turning point $P$ - Edexcel - A-Level Maths Pure - Question 8 - 2009 - Paper 2

Find the exact coordinates of P.

Use the iterative formula with x_0 = 0.25 to find, to 4 decimal places, the values of x_1, x_2, and x_3.

By choosing a suitable interval, show that a root of f(x) = 0 is x = 0.2576 correct to 4 decimal places.

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