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

The curve C has equation $y = f(x)$, $x > 0$, where $f'(x) = 30 + \frac{6 - 5x^2}{\sqrt{x}}$ - Edexcel - A-Level Maths Pure - Question 9 - 2017 - Paper 1

Question 9

$The-curve-C-has-equation-$y-=-f(x)$,-$x->-0$,-where--$f'(x)-=-30-+-\frac{6---5x^2}{\sqrt{x}}$-Edexcel-A-Level Maths Pure-Question 9-2017-Paper 1.png$

The curve C has equation $y = f(x)$, $x > 0$, where $f'(x) = 30 + \frac{6 - 5x^2}{\sqrt{x}}$. Given that the point $P(4, -8)$ lies on C, a) find the equation of t... show full transcript

Worked Solution & Example Answer:The curve C has equation $y = f(x)$, $x > 0$, where $f'(x) = 30 + \frac{6 - 5x^2}{\sqrt{x}}$ - Edexcel - A-Level Maths Pure - Question 9 - 2017 - Paper 1

Step 1

find the equation of the tangent to C at P, giving your answer in the form y = mx + c, where m and c are constants.

96%

114 rated

Answer

To find the equation of the tangent at point P(4, -8), we first need to find the derivative of the function at $x = 4$ :

Substitute $x = 4$ into the derivative:

$f'(4) = 30 + \frac{6 - 5(4^2)}{\sqrt{4}}$ $f'(4) = 30 + \frac{6 - 80}{2}$ $f'(4) = 30 - 37 = -7$
The gradient (m) of the tangent line at P is -7. Thus, we can use the point-slope form of the equation:

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

Where $(x_1, y_1) = (4, -8)$ and $m = -7$ :

$y - (-8) = -7(x - 4)$ $y + 8 = -7x + 28$ $y = -7x + 20$

Thus, the equation of the tangent line is:

$y = -7x + 20$

Step 2

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

99%

104 rated

Answer

To find $f(x)$ , we need to integrate the first derivative:

$f'(x) = 30 + \frac{6 - 5x^2}{\sqrt{x}}$

Let's break this down:

$f'(x) = 30 + 6x^{-1/2} - 5x^{3/2}$

Now integrate term by term:

$f(x) = \int (30) \, dx + \int (6x^{-1/2}) \, dx - \int (5x^{3/2}) \, dx$

Calculating each integral separately, we get:

$\int 30 \, dx = 30x$
$\int 6x^{-1/2} \, dx = 6 \cdot 2x^{1/2} = 12\sqrt{x}$
$\int -5x^{3/2} \, dx = -5 \cdot \frac{2}{5} x^{5/2} = -2x^{5/2}$

Combining these results, we have:

$f(x) = 30x + 12\sqrt{x} - 2x^{5/2} + C$

Where $C$ is the constant of integration.

The curve C has equation $y = f(x)$, $x > 0$, where $f'(x) = 30 + \frac{6 - 5x^2}{\sqrt{x}}$ - Edexcel - A-Level Maths Pure - Question 9 - 2017 - Paper 1

Worked Solution & Example Answer:The curve C has equation $y = f(x)$, $x > 0$, where $f'(x) = 30 + \frac{6 - 5x^2}{\sqrt{x}}$ - Edexcel - A-Level Maths Pure - Question 9 - 2017 - Paper 1

find the equation of the tangent to C at P, giving your answer in the form y = mx + c, where m and c are constants.

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

Join the A-Level students using SimpleStudy...