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Figure 8 shows a sketch of the curve C with equation \( y = x^{\frac{1}{3}}, x > 0 \) (a) Find, by firstly taking logarithms, the \( x \) coordinate of the turning point of C - Edexcel - A-Level Maths Pure - Question 11 - 2019 - Paper 2
Question 11
Figure 8 shows a sketch of the curve C with equation \( y = x^{\frac{1}{3}}, x > 0 \) (a) Find, by firstly taking logarithms, the \( x \) coordinate of the turning ... show full transcript
Worked Solution & Example Answer:Figure 8 shows a sketch of the curve C with equation \( y = x^{\frac{1}{3}}, x > 0 \) (a) Find, by firstly taking logarithms, the \( x \) coordinate of the turning point of C - Edexcel - A-Level Maths Pure - Question 11 - 2019 - Paper 2
Step 1
Find, by firstly taking logarithms, the \( x \) coordinate of the turning point of C.
Answer
To find the turning point of the curve ( y = x^{\frac{1}{3}} ), we first take the natural logarithm of both sides:
Differentiating both sides with respect to ( x ) gives:
Setting ( \frac{dy}{dx} = 0 ) allows us to find critical points:
Next, we take the derivative of ( \log y ) with respect to ( x ):
To find the value of ( x ), we set this equation to zero:
Step 2
Show that \( 1.5 < \alpha < 1.6 \).
Answer
To show that ( 1.5 < \alpha < 1.6 ), we verify the function values at these points:
- Calculate ( f(1.5) = 1.5^{\frac{1}{3}} \approx 1.1447 ) and ( f(1.6) = 1.6^{\frac{1}{3}} \approx 1.1680.)
- Since point P(2, 2) lies on C, we find that at these points the curve exhibits continuity. Thus, the appropriate range is confirmed:
Since the curve transitions from below to above 2, we conclude:
Step 3
Using this formula with \( x_1 = 1.5 \), find \( x_3 \) to 3 decimal places.
Answer
Starting with the iteration formula:
x_{n+1} = 2x_n - x_n^{\frac{1}{3}}\n$$ Plugging in our initial value: 1. \( x_1 = 1.5 \) - \( x_2 = 2 \cdot 1.5 - 1.5^{\frac{1}{3}} \approx 2.124 - 1.1447 \approx 1.9793 \) 2. \( x_2 \approx 1.9793 \) - \( x_3 = 2 \cdot 1.9793 - 1.9793^{\frac{1}{3}} \approx 3.9586 - 1.2682 \approx 2.6904. \) Thus, rounding gives: \( x_3 \approx 2.690.\)Step 4
Describe the long-term behaviour of \( x_n \).
Answer
The iteration formula suggests a behaviour that converges through oscillation toward the limit. As ( n ) increases, the value of ( x_n ) approaches the root of the equation. Given that the sequence generated approaches a stable limit and oscillates between two values before converging, we conclude that:
The behaviour indicates convergence, specifically towards ( \alpha ) around 1.7. Thus, the sequence stabilizes over time.