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

Question 7

The curve with equation $y = f(x) = 3x e^x - 1$ 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 7 - 2009 - Paper 2

Step 1

Find the exact coordinates of P.

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Answer

To find the coordinates of the turning point $P$ , we first need to determine where the first derivative $f'(x)$ is equal to zero.

The function is given by: $f(x) = 3x e^x - 1$

Calculating the derivative: $f'(x) = 3 e^x + 3x e^x = 3 e^x (1 + x)$

Setting the derivative to zero for turning points: $3 e^x (1 + x) = 0$

Here, $e^x$ is never zero, hence: $1 + x = 0 \Rightarrow x = -1$

Substituting back into $f(x)$ to find the $y$ -coordinate: $f(-1) = 3(-1)e^{-1} - 1 = -\frac{3}{e} - 1$

Thus, the exact coordinates of $P$ are: $$P = \left(-1, -\frac{3}{e} - 1\right)$.

Step 2

Use the iterative formula with $x_0 = 0.25$ to find $x_1$, $x_2$, and $x_3$.

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Answer

We will use the iterative formula: $x_{n+1} = \frac{1}{3} e^{x_n}$

Starting with $x_0 = 0.25$ :

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

Thus, the values are:

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

Step 3

Show that a root of $f(x) = 0$ is $x = 0.2576$ correct to 4 decimal places.

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Answer

To confirm a root exists near $0.2576$ , we evaluate the function $f(x)$ at suitable intervals:

Choosing $(0.2575, 0.25765)$ or similar narrow intervals:

Evaluate:
- $f(0.2575)$
- $f(0.25765)$

Using these evaluations:

$f(0.2575) \approx 0.000379$ (a positive value)
$f(0.25765) \approx 0.000109$ (a smaller positive value)

Since $f(0.2576)$ is approximately $0$ , it can be stated that a root in the interval $(0.2575, 0.25765)$ indicates that $x = 0.2576$ 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 7 - 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 7 - 2009 - Paper 2

Find the exact coordinates of P.

Use the iterative formula with $x_0 = 0.25$ to find $x_1$, $x_2$, and $x_3$.

Show that a root of $f(x) = 0$ is $x = 0.2576$ correct to 4 decimal places.

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