A curve has equation y = f(x). The point P with coordinates (9, 0) lies on the curve. Given that $$f'(x) = \frac{x + 9}{\sqrt{x}}, \, x > 0$$ (a) find f(x). (b) Find the x-coordinates of the two points on y = f(x) where the gradient of the curve is equal to 10.

A-Level Edexcel Maths Pure Exponential & Logarithms: A curve has equation y = f(x). T

A curve has equation y = f(x) - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 2

Question 5

A curve has equation y = f(x). The point P with coordinates (9, 0) lies on the curve. Given that $$f'(x) = \frac{x + 9}{\sqrt{x}}, \, x > 0$$ (a) find f(x). (b) ... show full transcript

Worked Solution & Example Answer:A curve has equation y = f(x) - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 2

Step 1

find f(x).

96%

114 rated

Answer

To find the function f(x) from its derivative f'(x), we need to integrate the expression:

$f(x) = \int f'(x) \, dx = \int \frac{x + 9}{\sqrt{x}} \, dx$

We can split the integral into two parts:

$\int \frac{x}{\sqrt{x}} \, dx + \int \frac{9}{\sqrt{x}} \, dx$

This simplifies to:

$\int x^{1/2} \, dx + 9 \int x^{-1/2} \, dx$

Integrating each term gives:

$= \frac{2}{3}x^{3/2} + 9(2x^{1/2}) + C = \frac{2}{3}x^{3/2} + 18\sqrt{x} + C$

Now, we need to find the constant C. Since (9, 0) lies on the curve:

ightarrow \frac{2}{3}(9^{3/2}) + 18(\sqrt{9}) + C = 0$$ Calculating: $$\frac{2}{3}(27) + 54 + C = 0$$ Thus, $$18 + C = 0 ightarrow C = -18$$ Therefore, the function is: $$f(x) = \frac{2}{3}x^{3/2} + 18\sqrt{x} - 18$$

Step 2

Find the x-coordinates of the two points on y = f(x) where the gradient of the curve is equal to 10.

99%

104 rated

Answer

We need to set the derivative equal to 10:

$f'(x) = \frac{x + 9}{\sqrt{x}} = 10$

Multiplying both sides by (\sqrt{x}) gives:

$x + 9 = 10\sqrt{x}$

Rearranging gives:

$x - 10\sqrt{x} + 9 = 0$

Let (y = \sqrt{x}) so we can rewrite the equation as:

$y^2 - 10y + 9 = 0$

Factoring:

$(y - 1)(y - 9) = 0$

Thus, we find:

$y = 1 \, ext{or} \, y = 9$

Going back to x:

$\sqrt{x} = 1 \Rightarrow x = 1$ $\sqrt{x} = 9 \Rightarrow x = 81$

Therefore, the x-coordinates of the points are 1 and 81.

A curve has equation y = f(x) - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 2

Worked Solution & Example Answer:A curve has equation y = f(x) - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 2

find f(x).

Find the x-coordinates of the two points on y = f(x) where the gradient of the curve is equal to 10.

Join the A-Level students using SimpleStudy...