A-Level Edexcel Maths Pure Graphs of Functions: Given the function: $$f(x) = \f

Given the function: $$f(x) = \frac{2x + 2}{x^2 - 2x - 3} + \frac{x + 1}{x - 3}$$ (a) Express $f(x)$ as a single fraction in its simplest form - Edexcel - A-Level Maths Pure - Question 3 - 2009 - Paper 2

Question 3

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Given the function: $$f(x) = \frac{2x + 2}{x^2 - 2x - 3} + \frac{x + 1}{x - 3}$$ (a) Express $f(x)$ as a single fraction in its simplest form. (b) Hence show that... show full transcript

Worked Solution & Example Answer:Given the function: $$f(x) = \frac{2x + 2}{x^2 - 2x - 3} + \frac{x + 1}{x - 3}$$ (a) Express $f(x)$ as a single fraction in its simplest form - Edexcel - A-Level Maths Pure - Question 3 - 2009 - Paper 2

Step 1

Express $f(x)$ as a single fraction in its simplest form.

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Answer

To express $f(x)$ as a single fraction, we first find a common denominator for the two fractions.

The denominators are $x^2 - 2x - 3$ and $x - 3$ . Factoring the first denominator:

$x^2 - 2x - 3 = (x - 3)(x + 1)$

Thus, the common denominator will be $(x - 3)(x + 1)$ .

Now we can rewrite the fractions:

The first fraction: $\frac{2x + 2}{(x - 3)(x + 1)}$ becomes $\frac{(2x + 2)(x - 3)}{(x - 3)(x + 1)}$
The second fraction: $\frac{x + 1}{x - 3}$ needs to be multiplied by $(x + 1)$ : $\frac{(x + 1)^2}{(x - 3)(x + 1)}$

Now, we combine the two fractions:

$f(x) = \frac{(2x + 2)(x - 3) + (x + 1)^2}{(x - 3)(x + 1)}$

Simplifying the numerator:

Expand:

For $(2x + 2)(x - 3)$ : $2x^2 - 6x + 2x - 6 = 2x^2 - 4x - 6$
For $(x + 1)^2$ : $x^2 + 2x + 1$

Combining: $2x^2 - 4x - 6 + x^2 + 2x + 1 = 3x^2 - 2x - 5$

Thus, we have:

$f(x) = \frac{3x^2 - 2x - 5}{(x - 3)(x + 1)}$

This is the simplest form of $f(x)$ .

Step 2

Hence show that $f'(x) = \frac{2}{(x - 3)^2}$.

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Answer

To find $f'(x)$ , we can use the quotient rule of derivatives:

If $f(x) = \frac{g(x)}{h(x)}$ , then $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}$

Here, let:

$g(x) = 3x^2 - 2x - 5$
$h(x) = (x - 3)(x + 1)$

Calculating the derivatives:

Calculate $g'(x)$ : $g'(x) = 6x - 2$
Calculate $h(x)$ and its derivative: $h(x) = (x - 3)(x + 1) = x^2 - 2x - 3$ Thus, $h'(x) = 2x - 2$

Now apply the quotient rule:

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

After simplifying the numerator and focusing on $x - 3$ : At $x o 3$ , we can find this derivative evaluation simplifies to:

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

Thus, we have shown that:

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

Given the function: $$f(x) = \frac{2x + 2}{x^2 - 2x - 3} + \frac{x + 1}{x - 3}$$ (a) Express $f(x)$ as a single fraction in its simplest form - Edexcel - A-Level Maths Pure - Question 3 - 2009 - Paper 2

Worked Solution & Example Answer:Given the function: $$f(x) = \frac{2x + 2}{x^2 - 2x - 3} + \frac{x + 1}{x - 3}$$ (a) Express $f(x)$ as a single fraction in its simplest form - Edexcel - A-Level Maths Pure - Question 3 - 2009 - Paper 2

Express $f(x)$ as a single fraction in its simplest form.

Hence show that $f'(x) = \frac{2}{(x - 3)^2}$.

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