The curve with equation $y = f(x) = 3xe^x - 1$ 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 $x_3$. (c) 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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The curve with equation $y = f(x) = 3xe^x - 1$ has a turning point $P$ - Edexcel - A-Level Maths Pure - Question 7 - 2009 - Paper 2

Question 7

The curve with equation $y = f(x) = 3xe^x - 1$ 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 $y = f(x) = 3xe^x - 1$ has a turning point $P$ - Edexcel - A-Level Maths Pure - Question 7 - 2009 - Paper 2

Step 1

Find the exact coordinates of P.

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Answer

To find the exact coordinates of the turning point $P$ , we start by finding the first derivative of the function:

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

Setting the derivative to zero to find critical points:

$3e^x(1 + x) = 0$
Since $e^x$ is never zero, we solve:

$1 + x = 0 \Rightarrow x = -1$

To find the $y$ -coordinate of $P$ , we substitute $x = -1$ back into the original function:

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

Thus, the coordinates of the turning point $P$ are $ig(-1, -\frac{3}{e} - 1\big)$ .

Step 2

Use the iterative formula

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Answer

Using the iterative formula:

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

Calculate $x_1$ :
- Set $x_0 = 0.25$ .
- Calculate: $x_1 = \frac{1}{3} e^{0.25} \approx 0.2596$
Calculate $x_2$ :
- Set $x_1 = 0.2596$ .
- Calculate: $x_2 = \frac{1}{3} e^{0.2596} \approx 0.2571$
Calculate $x_3$ :
- Set $x_2 = 0.2571$ .
- Calculate: $x_3 = \frac{1}{3} e^{0.2571} \approx 0.2578$

Thus, the values are approximately:

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

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 there is a root near $x = 0.2576$ , we can examine the values of $f(x)$ in the interval (0.2575, 0.25765).

Calculate $f(0.2575) ext{ and } f(0.25765)$ $f (0.2575) e x t an d f (0.25765)$ :
- $f(0.2575) \approx 0.000379$
- $f(0.25765) \approx -0.000109$

Since $f(0.2575)$ is positive and $f(0.25765)$ is negative, by the Intermediate Value Theorem, there exists at least one root in the interval $(0.2575, 0.25765)$ . Therefore, we conclude:

$x = 0.2576$

is correct to 4 decimal places.

The curve with equation $y = f(x) = 3xe^x - 1$ has a turning point $P$ - Edexcel - A-Level Maths Pure - Question 7 - 2009 - Paper 2

Worked Solution & Example Answer:The curve with equation $y = f(x) = 3xe^x - 1$ has a turning point $P$ - Edexcel - A-Level Maths Pure - Question 7 - 2009 - Paper 2

Find the exact coordinates of P.

Use the iterative formula

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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