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Given that $k$ is a negative constant and that the function $f(x)$ is defined by $$f(x) = 2 - \frac{(x - 5k)(x - k)}{x^2 - 3kx + 2k^2}, \quad x \geq 0$$ (a) show that $f(x) = \frac{x + k}{x - 2k}$ (b) Hence find $f' (x)$, giving your answer in its simplest form - Edexcel - A-Level Maths Pure - Question 2 - 2014 - Paper 5
Question 2
Given that $k$ is a negative constant and that the function $f(x)$ is defined by $$f(x) = 2 - \frac{(x - 5k)(x - k)}{x^2 - 3kx + 2k^2}, \quad x \geq 0$$ (a) show t... show full transcript
Worked Solution & Example Answer:Given that $k$ is a negative constant and that the function $f(x)$ is defined by $$f(x) = 2 - \frac{(x - 5k)(x - k)}{x^2 - 3kx + 2k^2}, \quad x \geq 0$$ (a) show that $f(x) = \frac{x + k}{x - 2k}$ (b) Hence find $f' (x)$, giving your answer in its simplest form - Edexcel - A-Level Maths Pure - Question 2 - 2014 - Paper 5
Step 1
show that $f(x) = \frac{x + k}{x - 2k}$
Answer
To determine whether ( f(x) ) is increasing or decreasing, we examine the sign of ( f'(x) ).
We have:
Since ( k ) is negative, ( -3k ) is positive. The denominator ( (x - 2k)^2 ) is always positive for ( x \geq 0 ). Therefore, ( f'(x) > 0 ) for all valid values of ( x ).
This indicates that ( f(x) ) is an increasing function over the specified domain.