A curve with equation y = f(x) passes through the point (4, 9). Given that f'(x) = \frac{3\sqrt{x}}{2} - \frac{9}{4\sqrt{x}} + 2, \quad x > 0 a) find f(x), giving each term in its simplest form. Point P lies on the curve. The normal to the curve at P is parallel to the line 2y + x = 0. b) Find the x coordinate of P.

A-Level Edexcel Maths Pure Integration: A curve with equation y = f(x) p

A curve with equation y = f(x) passes through the point (4, 9) - Edexcel - A-Level Maths Pure - Question 3 - 2015 - Paper 1

Question 3

A curve with equation y = f(x) passes through the point (4, 9). Given that f'(x) = \frac{3\sqrt{x}}{2} - \frac{9}{4\sqrt{x}} + 2, \quad x > 0 a) find f(x), giving... show full transcript

Worked Solution & Example Answer:A curve with equation y = f(x) passes through the point (4, 9) - Edexcel - A-Level Maths Pure - Question 3 - 2015 - Paper 1

Step 1

a) find f(x), giving each term in its simplest form.

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Answer

To find the function f(x) from its derivative f'(x), we will integrate f'(x).

Starting with:

$f'(x) = \frac{3\sqrt{x}}{2} - \frac{9}{4\sqrt{x}} + 2$

We can rewrite this as:

$f'(x) = \frac{3}{2} x^{1/2} - \frac{9}{4} x^{-1/2} + 2$

Now, integrating each term separately:

$f(x) = \int \left( \frac{3}{2} x^{1/2} \right) dx - \int \left( \frac{9}{4} x^{-1/2} \right) dx + \int 2 dx$

Evaluating the integrals:

$\int \frac{3}{2} x^{1/2} dx = \frac{3}{2} \cdot \frac{2}{3} x^{3/2} = x^{3/2}$
$\int \frac{9}{4} x^{-1/2} dx = \frac{9}{4} \cdot 2 x^{1/2} = \frac{9}{2} x^{1/2}$
$\int 2 dx = 2x$

Thus, combining these results,

$f(x) = x^{3/2} - \frac{9}{2} x^{1/2} + 2x + C$

To find the constant C, we use the point (4, 9):

$f(4) = 4^{3/2} - \frac{9}{2} \cdot 4^{1/2} + 2 \cdot 4 + C = 9$

Calculating the left side:

$4^{3/2} = 8$
$\frac{9}{2} \cdot 2 = 9$
$2 \cdot 4 = 8$

Therefore,

$8 - 9 + 8 + C = 9$ $7 + C = 9$ $C = 2$

Hence, we have:

$f(x) = x^{3/2} - \frac{9}{2} x^{1/2} + 2x + 2$

Step 2

b) Find the x coordinate of P.

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Answer

We know that the normal to the curve at point P is parallel to the line 2y + x = 0.

First, find the slope of this line:

From the equation, we can express y:

$2y = -x \Rightarrow y = -\frac{1}{2} x$

So, the slope (m_normal) of the normal is:

$m_{normal} = -\frac{1}{2}$

The slope of the tangent line (m_tangent) is the negative reciprocal of the slope of the normal:

$m_{tangent} = - \frac{1}{m_{normal}} = 2$

Now, we will find where the derivative f'(x) equals 2:

$\frac{3\sqrt{x}}{2} - \frac{9}{4\sqrt{x}} + 2 = 2$

Subtracting 2 from both sides gives:

$\frac{3\sqrt{x}}{2} - \frac{9}{4\sqrt{x}} = 0$

Multiplying through by 4\sqrt{x} to eliminate the fractions yields:

$6x - 9 = 0 \Rightarrow 6x = 9 \Rightarrow x = \frac{3}{2}$

Thus, the x coordinate of point P is:

$x = \frac{3}{2}$

A curve with equation y = f(x) passes through the point (4, 9) - Edexcel - A-Level Maths Pure - Question 3 - 2015 - Paper 1

Worked Solution & Example Answer:A curve with equation y = f(x) passes through the point (4, 9) - Edexcel - A-Level Maths Pure - Question 3 - 2015 - Paper 1

a) find f(x), giving each term in its simplest form.

b) Find the x coordinate of P.

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