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Figure 1 shows a sketch of the curve C with equation y = \frac{4x^{2}+x}{2\sqrt{x}} \quad x > 0 (a) Show that dy \over dx = \frac{12x^{2}+x-16\sqrt{x}}{4\sqrt{x}}\n\nThe point P, shown in Figure 1, is the minimum turning point on C - Edexcel - A-Level Maths Pure - Question 7 - 2020 - Paper 2
Question 7
Figure 1 shows a sketch of the curve C with equation y = \frac{4x^{2}+x}{2\sqrt{x}} \quad x > 0 (a) Show that dy \over dx = \frac{12x^{2}+x-16\sqrt{x}}{4\sqrt{x}}... show full transcript
Worked Solution & Example Answer:Figure 1 shows a sketch of the curve C with equation y = \frac{4x^{2}+x}{2\sqrt{x}} \quad x > 0 (a) Show that dy \over dx = \frac{12x^{2}+x-16\sqrt{x}}{4\sqrt{x}}\n\nThe point P, shown in Figure 1, is the minimum turning point on C - Edexcel - A-Level Maths Pure - Question 7 - 2020 - Paper 2
Step 1
Show that dy/dx = \frac{12x^{2}+x-16\sqrt{x}}{4\sqrt{x}}
Answer
To find ( \frac{dy}{dx} ), we differentiate the function:
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Start with the given equation: [ y = \frac{4x^{2}+x}{2\sqrt{x}} ]
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Use the quotient rule: [ \frac{dy}{dx} = \frac{(u/v)'} = \frac{v(u') - u(v')}{v^{2}} ] where ( u = 4x^{2} + x ) and ( v = 2\sqrt{x} ).
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Calculate ( u' ) and ( v' ): ( u' = 8x + 1 ) ( v = 2x^{1/2} ) implies ( v' = x^{-1/2} )
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Substitute into the quotient rule: [ \frac{dy}{dx} = \frac{2\sqrt{x}(8x+1) - (4x^{2}+x)(\frac{1}{\sqrt{x}})}{(2\sqrt{x})^{2}} ]
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Simplify: [ \frac{dy}{dx} = \frac{12x^{2}+x-16\sqrt{x}}{4\sqrt{x}} ] and this confirms the result.
Step 2
Show that the x coordinate of P is a solution of x = (\frac{4 - \sqrt{3}}{3})^{\frac{3}{2}}
Answer
To find the x coordinate of P:
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Set ( \frac{dy}{dx} = 0 ) to find critical points: [ 12x^{2} + x - 16\sqrt{x} = 0 ]
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Rearranging gives: [ 12x^{2} + x - 16x^{1/2} = 0 ]
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Let ( z = \sqrt{x} ), then ( x = z^{2} ), substitute: Here we have: [ 12z^{4} + z - 16z = 0 ]
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This solves down to: [ z^{2} = \sqrt{\frac{4 - \sqrt{3}}{3}} \quad \text{thus } \quad x = \left( \frac{4 - \sqrt{3}}{3} \right)^{\frac{3}{2}} ] confirming the coordinate of point P.
Step 3
Use the iteration formula x_{n+1} = (\frac{4 - \sqrt{3}}{3})^{\frac{3}{2}} with x_{1} = 2 to find (i) the value of x_{2} to 5 decimal places.
Answer
To use the iteration formula:
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Start with ( x_{1} = 2 ).
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Apply the formula: [ x_{2} = \left( \frac{4 - \sqrt{3}}{3} \right)^{\frac{3}{2}} \approx 1.1656 ]
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Converge to five decimal places and you get: [ x_{2} = 1.16565 ]
Step 4