A-Level Edexcel Maths Pure Applications of Differentiation: The curve C has equation y = f(x

The curve C has equation y = f(x), x > 0, and f'(x) = 4 - 6 adix{x} + rac{8}{x^2} - Edexcel - A-Level Maths Pure - Question 10 - 2008 - Paper 2

Question 10

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The curve C has equation y = f(x), x > 0, and f'(x) = 4 - 6 adix{x} + rac{8}{x^2}. Given that the point P(4, 1) lies on C, (a) find f(x) and simplify your answer.... show full transcript

Worked Solution & Example Answer:The curve C has equation y = f(x), x > 0, and f'(x) = 4 - 6 adix{x} + rac{8}{x^2} - Edexcel - A-Level Maths Pure - Question 10 - 2008 - Paper 2

Step 1

find f(x) and simplify your answer.

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Answer

To find f(x), we need to integrate f'(x).

Given:

adix{x} + rac{8}{x^2}$$ We will integrate term-by-term: 1. The integral of 4 is: $$\int 4 \,dx = 4x$$ 2. The integral of $-6\radix{x}$ (which is $-6x^{\frac{1}{2}}$) is: $$\int -6x^{\frac{1}{2}} \,dx = -6 \cdot \frac{2}{3} x^{\frac{3}{2}} = -4x^{\frac{3}{2}}$$ 3. The integral of $\frac{8}{x^2}$ (which is $8x^{-2}$) is: $$\int 8x^{-2} \,dx = 8 \cdot -x^{-1} = -\frac{8}{x}$$ So, $$f(x) = 4x - 4x^{\frac{3}{2}} - \frac{8}{x} + C$$ Now, to simplify this, we can find the constant C using the condition P(4, 1) lies on curve C: $$f(4) = 1$$ Substituting: $$1 = 4(4) - 4(4^{\frac{3}{2}}) - \frac{8}{4} + C$$ This simplifies to: $$1 = 16 - 32 - 2 + C$$ Thus: $$C = 1 + 16 + 2 = 19$$ Therefore, $$f(x) = 4x - 4x^{\frac{3}{2}} - \frac{8}{x} + 19$$.

Step 2

Find an equation of the normal to C at the point P(4, 1).

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Answer

To find the equation of the normal to the curve at point P(4, 1), we first need to determine the slope of the tangent line at that point using the derivative:

From earlier, we have: $f'(x) = 4 - 6\radix{x} + \frac{8}{x^2}$

Substituting $x = 4$ : $f'(4) = 4 - 6\radix{4} + \frac{8}{4^2}$

Calculate: $f'(4) = 4 - 6(2) + \frac{8}{16} = 4 - 12 + 0.5 = -7.5$

The slope of the normal line is the negative reciprocal of the slope of the tangent: $m_{normal} = -\frac{1}{-7.5} = \frac{2}{15}$

Now, we can use point-slope form to write the equation of the normal: $y - y_1 = m(x - x_1)$ $y - 1 = \frac{2}{15}(x - 4)$

Hence, the equation becomes: $y - 1 = \frac{2}{15}x - \frac{8}{15}$ $y = \frac{2}{15}x - \frac{8}{15} + 1$ $y = \frac{2}{15}x + \frac{7}{15}$ .

The curve C has equation y = f(x), x > 0, and f'(x) = 4 - 6 adix{x} + rac{8}{x^2} - Edexcel - A-Level Maths Pure - Question 10 - 2008 - Paper 2

Worked Solution & Example Answer:The curve C has equation y = f(x), x > 0, and f'(x) = 4 - 6 adix{x} + rac{8}{x^2} - Edexcel - A-Level Maths Pure - Question 10 - 2008 - Paper 2

find f(x) and simplify your answer.

Find an equation of the normal to C at the point P(4, 1).

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