The curve C has equation $y=f(x)$, $x \neq 0$, and the point P(2, 1) lies on C. Given that $$f'(x) = 3x^2 - 6 - \frac{8}{x^2},$$ (a) find $f(x)$. (b) Find an equation for the tangent to C at the point P, giving your answer in the form $y=mx+c$, where m and c are integers.

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

The curve C has equation $y=f(x)$, $x \neq 0$, and the point P(2, 1) lies on C - Edexcel - A-Level Maths Pure - Question 8 - 2007 - Paper 2

Question 8

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The curve C has equation $y=f(x)$, $x \neq 0$, and the point P(2, 1) lies on C. Given that $$f'(x) = 3x^2 - 6 - \frac{8}{x^2},$$ (a) find $f(x)$. (b) Find an equati... show full transcript

Worked Solution & Example Answer:The curve C has equation $y=f(x)$, $x \neq 0$, and the point P(2, 1) lies on C - Edexcel - A-Level Maths Pure - Question 8 - 2007 - Paper 2

Step 1

find $f(x)$

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Answer

To find $f(x)$ , we start by integrating $f'(x)$ :

$f'(x) = 3x^2 - 6 - \frac{8}{x^2}$
Now we integrate term by term:

The integral of $3x^2$ is $x^3$
The integral of $-6$ is $-6x$
The integral of $-\frac{8}{x^2}$ is $8x^{-1}$ , which can be written as $-\frac{8}{x}$

Thus, we have:

$f(x) = x^3 - 6x + \frac{8}{x} + C$

Next, we use the fact that the point P(2, 1) lies on the curve to find the constant $C$ .

Substituting $x = 2$ and $f(2) = 1$ into the equation:
$1 = (2)^3 - 6(2) + \frac{8}{2} + C$
This simplifies to:
$1 = 8 - 12 + 4 + C$
$1 = 0 + C$
Thus, $C = 1$ . Therefore, the final function is:

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

Step 2

Find an equation for the tangent to C at the point P

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Answer

To find the equation of the tangent line at the point P(2, 1), we need to determine the slope at this point. The slope of the tangent line is given by $f'(2)$ :

Substituting $x = 2$ into $f'(x)$ : $f'(2) = 3(2)^2 - 6 - \frac{8}{(2)^2}$
Calculating this, we have: $f'(2) = 3(4) - 6 - \frac{8}{4} = 12 - 6 - 2 = 4$
So, the slope $m = 4$ .

Next, we can use the point-slope form of a line to write the equation of the tangent: $y - y_1 = m(x - x_1)$
Substituting $m = 4$ , $x_1 = 2$ , and $y_1 = 1$ gives us: $y - 1 = 4(x - 2)$
Expanding this, we find: $y - 1 = 4x - 8$
$y = 4x - 7$
Thus, the equation of the tangent line is:

$y = 4x - 7$

The curve C has equation $y=f(x)$, $x \neq 0$, and the point P(2, 1) lies on C - Edexcel - A-Level Maths Pure - Question 8 - 2007 - Paper 2

Worked Solution & Example Answer:The curve C has equation $y=f(x)$, $x \neq 0$, and the point P(2, 1) lies on C - Edexcel - A-Level Maths Pure - Question 8 - 2007 - Paper 2

find $f(x)$

Find an equation for the tangent to C at the point P

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