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Figure 8 shows a sketch of the curve C with equation $y = x^{ rac{1}{3}}$, $x > 0$ - Edexcel - A-Level Maths Pure - Question 12 - 2019 - Paper 2
Question 12
Figure 8 shows a sketch of the curve C with equation $y = x^{ rac{1}{3}}$, $x > 0$. (a) Find, by firstly taking logarithms, the $x$ coordinate of the turning point... show full transcript
Worked Solution & Example Answer:Figure 8 shows a sketch of the curve C with equation $y = x^{ rac{1}{3}}$, $x > 0$ - 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.
Answer
I will analyze the given mathematical expressions and set them to find the critical points such that the curve has no slope, thus representing the shape of the curve, but directly translating through logarithmic methods we observe that:
The x-coordinate of the turning point is
Step 2
Show that $1.5 < a < 1.6$.
Answer
To show that the point lies on C, we substitute into the equation
\ ext{Rearranging gives us:} \\ \ a = 2^3 = 8. \ \ ext{To find bounds for } a, we consider: } \\ y = x^{ rac{1}{3}}\ \ ext{Setting 1.5 for } a:\\ 1.5^{ rac{1}{3}} \ ext{and 1.6:} \ \ ext{Calculating, we see: } \ 1.5^{ rac{1}{3}} ext{ gives aproximately } 1.1449 \ ext{ and similarly } \ 1.6^{ rac{1}{3}} ext{ is roughly } 1.1803. \ ext{So, we observe that } 1.5 < a < 1.6.\ \Step 3
Step 4
describe the long-term behaviour of $x_n$.
Answer
As approaches infinity, the iterative process shows that converges towards a fixed point.
In this setup, provided the iteration remains bounded, all sequences oscillate between two points and stabilize as they converge to this limit. Hence, we state:
- If persists oscillating through iterations, it may approach a decimal limit over time. Furthermore, it is indicative of convergence to an established limit or oscillatory pattern. This is supported by iteratively confirming convergence.