Figure 8 shows a sketch of the curve C with equation $y = x^x$, $x > 0$. (a) Find, by firstly taking logarithms, the x coordinate of the turning point of C. (Solutions based entirely on graphical or numerical methods are not acceptable.) The point P(α, 2) lies on C. (b) Show that 1.5 < α < 1.6. A possible iteration formula that could be used in an attempt to find α is $x_{n+1} = 2x_n^{x_n}$. Using this formula with $x_1 = 1.5$ (c) find $x_3$ to 3 decimal places, (d) describe the long-term behaviour of $x_n$.

A-Level Edexcel Maths Pure Absolute Value Equations: Figure 8 shows a sketch of the c

Figure 8 shows a sketch of the curve C with equation $y = x^x$, $x > 0$ - Edexcel - A-Level Maths Pure - Question 12 - 2019 - Paper 2

Question 12

Figure 8 shows a sketch of the curve C with equation $y = x^x$, $x > 0$. (a) Find, by firstly taking logarithms, the x coordinate of the turning point of C. (Solut... show full transcript

Worked Solution & Example Answer:Figure 8 shows a sketch of the curve C with equation $y = x^x$, $x > 0$ - Edexcel - A-Level Maths Pure - Question 12 - 2019 - Paper 2

Step 1

Find, by firstly taking logarithms, the x coordinate of the turning point of C.

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Answer

To find the turning point of the curve defined by the equation $y = x^x$ , we first take the natural logarithm of both sides:

ext{Taking logarithms:} & \ ext{Let } y = x^x \ ext{Then:} & \ ext{ln}(y) = x ext{ln}(x) ext{Differentiating both sides:} & \ \frac{dy}{dx} = rac{d}{dx}(x ext{ln}(x)) & \ \rightarrow \frac{dy}{dx} = \text{ln}(x) + 1 \\

To find the turning point, we set $rac{dy}{dx} = 0$ :

$ext{ln}(x) + 1 = 0$

Solving for $x$ gives us: $ext{ln}(x) = -1$

Thus the turning point occurs at $x \approx 0.368$ .

Step 2

Show that 1.5 < α < 1.6.

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Answer

Given the point P(α, 2) lies on the curve, we substitute this into the equation:

$2 = α^α$ .

Testing values between 1.5 and 1.6:

For $α = 1.5$ : $1.5^{1.5} \approx 1.8371,$

For $α = 1.6$ : $1.6^{1.6} \approx 1.7272.$

Since $1.8371 > 2$ and $1.7272 < 2$ , this confirms: $1.5 < α < 1.6.$

Step 3

Using this formula with $x_1 = 1.5$ find $x_3$ to 3 decimal places.

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Answer

Using the iteration formula $x_{n+1} = 2x_n^{x_n}$ with $x_1 = 1.5$ :

Calculate $x_2$ : $x_2 = 2(1.5)^{1.5} \approx 2 \cdot 1.8371 \approx 3.6742.$
Calculate $x_3$ : $x_3 = 2(3.6742)^{3.6742} \approx 2(51.4646) \approx 102.9292.$

Thus, $x_3 \approx 102.929.$

Step 4

Describe the long-term behaviour of $x_n$.

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Answer

The sequence defined by the iteration formula tends to diverge because as $x_n$ increases, the value of $2x_n^{x_n}$ grows rapidly. Thus, $x_n$ does not converge but will oscillate or diverge as observed from initial conditions, taking increasingly greater positive values.

Figure 8 shows a sketch of the curve C with equation $y = x^x$, $x > 0$ - Edexcel - A-Level Maths Pure - Question 12 - 2019 - Paper 2

Worked Solution & Example Answer:Figure 8 shows a sketch of the curve C with equation $y = x^x$, $x > 0$ - Edexcel - A-Level Maths Pure - Question 12 - 2019 - Paper 2

Find, by firstly taking logarithms, the x coordinate of the turning point of C.

Show that 1.5 < α < 1.6.

Using this formula with $x_1 = 1.5$ find $x_3$ to 3 decimal places.

Describe the long-term behaviour of $x_n$.

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