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 Equation of a Straight Line: The curve C has equation $y = f(

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

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

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 eq... 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 $f(x)$

96%

114 rated

Answer

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

$f'(x) = 3x^2 - 6 - \frac{8}{x^2}$

Integrating 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 $8x^{-1}$ or $-\frac{8}{x}$ .

Thus,

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

To find the constant $C$ , we use the point P(2, 1):

$1 = (2)^3 - 6(2) + \frac{8}{2} + C$ $1 = 8 - 12 + 4 + C$ $1 = 0 + C$ $C = 1$

Therefore, the equation for $f(x)$ is:

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

Step 2

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

99%

104 rated

Answer

At point P(2, 1), we first need to find the slope $m$ of the tangent line. We can find $m$ by evaluating $f'(x)$ at $x = 2$ :

$f'(2) = 3(2)^2 - 6 - \frac{8}{(2)^2}$ $f'(2) = 3(4) - 6 - \frac{8}{4}$ $f'(2) = 12 - 6 - 2 = 4$

Now that we have the slope $m = 4$ and the point P(2, 1), we can use the point-slope form of the line to find the equation:

$y - y_1 = m(x - x_1)$ $y - 1 = 4(x - 2)$

Simplifying gives:

$y - 1 = 4x - 8$ $y = 4x - 7$

Thus, the equation of the tangent at point P 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

Join the A-Level students using SimpleStudy...