The point P(4, -1) lies on the curve C with equation y = f(x), x > 0, and f'(x) = \frac{1}{2} - \frac{6}{\sqrt{x}} + 3. (a) Find the equation of the tangent to C at the point P, giving your answer in the form y = mx + c, where m and c are integers. (b) Find f(x).

A-Level Edexcel Maths Pure Graphs of Functions: The point P(4, -1) lies on the c

The point P(4, -1) lies on the curve C with equation y = f(x), x > 0, and f'(x) = \frac{1}{2} - \frac{6}{\sqrt{x}} + 3 - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 2

Question 8

$The-point-P(4,--1)-lies-on-the-curve-C-with-equation-y-=-f(x),-x->-0,--and--f'(x)-=-\frac{1}{2}---\frac{6}{\sqrt{x}}-+-3-Edexcel-A-Level Maths Pure-Question 8-2012-Paper 2.png$

The point P(4, -1) lies on the curve C with equation y = f(x), x > 0, and f'(x) = \frac{1}{2} - \frac{6}{\sqrt{x}} + 3. (a) Find the equation of the tangent to C ... show full transcript

Worked Solution & Example Answer:The point P(4, -1) lies on the curve C with equation y = f(x), x > 0, and f'(x) = \frac{1}{2} - \frac{6}{\sqrt{x}} + 3 - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 2

Step 1

Find the equation of the tangent to C at the point P

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Answer

To find the equation of the tangent at point P(4, -1), we first need to calculate f'(4):

$f'(4) = \frac{1}{2} - \frac{6}{\sqrt{4}} + 3 = \frac{1}{2} - 3 + 3 = \frac{1}{2}$

The slope (m) at point P is thus ( m = \frac{1}{2} ).

Using the point-slope form of a linear equation, we have:

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

Plugging in the values:

$y - (-1) = \frac{1}{2}(x - 4)$

This simplifies to:

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

Therefore:

$y = \frac{1}{2}x - 3$

We express this in the required form:

$y = \frac{1}{2}x - 3$
where m = 1/2 and c = -3.

Since m and c need to be integers, we multiply through by 2 to get:

$2y = x - 6$

This leads to the final answer for part (a):

$y = 1x - 6$ .

Thus, m = 1 and c = -6.

Step 2

Find f(x)

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Answer

To find f(x), we start with:

$f'(x) = \frac{1}{2} - \frac{6}{\sqrt{x}} + 3$ .

Integrating f'(x) to find f(x):

$f(x) = \int \left(\frac{1}{2} - \frac{6}{\sqrt{x}} + 3\right) dx$

Calculating term by term:

The integral of ( \frac{1}{2} ) is ( \frac{1}{2}x ).
The second term: ( \int -6x^{-\frac{1}{2}} dx = -6(2\sqrt{x}) = -12\sqrt{x} ).
The integral of 3 is ( 3x ).

Combining these gives:

$f(x) = \frac{1}{2}x - 12\sqrt{x} + 3x + c$

Thus:

$f(x) = \left(\frac{1}{2} + 3\right)x - 12\sqrt{x} + c = \frac{7}{2}x - 12\sqrt{x} + c$

To find c, we use the point P(4, -1):

$-1 = \frac{7}{2}(4) - 12\sqrt{4} + c$

This results in: $-1 = 14 - 24 + c$

Thus: $c = 9$

Putting it all together, the final expression for f(x) is:

$f(x) = \frac{7}{2}x - 12\sqrt{x} + 9$ .

The point P(4, -1) lies on the curve C with equation y = f(x), x > 0, and f'(x) = \frac{1}{2} - \frac{6}{\sqrt{x}} + 3 - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 2

Worked Solution & Example Answer:The point P(4, -1) lies on the curve C with equation y = f(x), x > 0, and f'(x) = \frac{1}{2} - \frac{6}{\sqrt{x}} + 3 - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 2

Find the equation of the tangent to C at the point P

Find f(x)

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