The curve C has equation $y = f(x), x > 0$, and $f'(x) = 4 - 6rac{x}{x^2} + rac{8}{x^3}$. Given that the point $P(4, 1)$ lies on C, (a) find $f(x)$ and simplify your answer. (b) Find an equation of the normal to C at the point $P(4, 1)$.

A-Level Edexcel Maths Pure Arithmetic Sequences & Series: The curve C has equation $y = f(

The curve C has equation $y = f(x), x > 0$, and $f'(x) = 4 - 6rac{x}{x^2} + rac{8}{x^3}$ - Edexcel - A-Level Maths Pure - Question 11 - 2008 - Paper 2

Question 11

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The curve C has equation $y = f(x), x > 0$, and $f'(x) = 4 - 6rac{x}{x^2} + rac{8}{x^3}$. Given that the point $P(4, 1)$ lies on C, (a) find $f(x)$ and simplify... show full transcript

Worked Solution & Example Answer:The curve C has equation $y = f(x), x > 0$, and $f'(x) = 4 - 6rac{x}{x^2} + rac{8}{x^3}$ - Edexcel - A-Level Maths Pure - Question 11 - 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 $f'(x)$ :

$f'(x) = 4 - 6 \cdot \frac{1}{x} + \frac{8}{x^3}$

Integrating each term, we get:

$f(x) = \int \left(4 - 6 \cdot \frac{1}{x} + \frac{8}{x^3}\right) dx$

Calculating the integrals:

$f(x) = 4x - 6\ln |x| - 4\cdot \frac{1}{x^2} + C$

To find the constant $C$ , we use the point $P(4, 1)$ :

$f(4) = 1 = 4(4) - 6\ln(4) - 4 \cdot \frac{1}{16} + C$

Solving this:

$1 = 16 - 6\ln(4) - \frac{1}{4} + C$

This simplifies to:

$C = 1 + 6\ln(4) - 15.75$

Letting $C = 3 + 6\ln(4)$ , we obtain:

$f(x) = 4x - 6\ln |x| - \frac{4}{x^2} + (3 + 6\ln(4))$

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 at point $P(4, 1)$ , we first need to calculate the derivative at that point:

$f'(4) = 4 - 6 \cdot \frac{1}{4} + \frac{8}{4^3}$

Calculating this yields:

$f'(4) = 4 - 1.5 + 0.125 = 2.625$

The gradient of the normal is the negative reciprocal of the derivative:

$m = -\frac{1}{f'(4)} = -\frac{1}{2.625} = -\frac{8}{21}$

Now we apply the point-slope form of the line equation:

$y - y_1 = m(x - x_1)$

Setting $(x_1, y_1) = (4, 1)$ , we have:

$y - 1 = -\frac{8}{21}(x - 4)$

This simplifies to:

$y = -\frac{8}{21}x + \frac{32}{21} + 1$

Thus, the equation of the normal is:

$y = -\frac{8}{21}x + \frac{53}{21}$

The curve C has equation $y = f(x), x > 0$, and $f'(x) = 4 - 6 rac{x}{x^2} + rac{8}{x^3}$ - Edexcel - A-Level Maths Pure - Question 11 - 2008 - Paper 2

Worked Solution & Example Answer:The curve C has equation $y = f(x), x > 0$, and $f'(x) = 4 - 6 rac{x}{x^2} + rac{8}{x^3}$ - Edexcel - A-Level Maths Pure - Question 11 - 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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The curve C has equation $y = f(x), x > 0$, and $f'(x) = 4 - 6rac{x}{x^2} + rac{8}{x^3}$ - Edexcel - A-Level Maths Pure - Question 11 - 2008 - Paper 2

Worked Solution & Example Answer:The curve C has equation $y = f(x), x > 0$, and $f'(x) = 4 - 6rac{x}{x^2} + rac{8}{x^3}$ - Edexcel - A-Level Maths Pure - Question 11 - 2008 - Paper 2