GCSE Edexcel Maths Probability: The functions f and g are such t

The functions f and g are such that f(x) = 3x - 1 and g(x) = x² + 4 (a) Find f⁻¹(x) Given that fg(x) = 2g(f(x)), (b) show that 15x² - 12x - 1 = 0. - Edexcel - GCSE Maths - Question 22 - 2019 - Paper 1

Question 22

The functions f and g are such that f(x) = 3x - 1 and g(x) = x² + 4 (a) Find f⁻¹(x) Given that fg(x) = 2g(f(x)), (b) show that 15x² - 12x - 1 = 0.

Worked Solution & Example Answer:The functions f and g are such that f(x) = 3x - 1 and g(x) = x² + 4 (a) Find f⁻¹(x) Given that fg(x) = 2g(f(x)), (b) show that 15x² - 12x - 1 = 0. - Edexcel - GCSE Maths - Question 22 - 2019 - Paper 1

Step 1

Find f⁻¹(x)

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Answer

To find the inverse function f⁻¹(x) of f(x) = 3x - 1, we need to solve for x in terms of y:

Start with the equation:
$y = 3x - 1$
Rearranging gives:
$y + 1 = 3x$
Solving for x yields:
$x = \frac{y + 1}{3}$
Therefore, the inverse function is:
$f^{-1}(x) = \frac{x + 1}{3}$

Step 2

show that 15x² - 12x - 1 = 0

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Answer

Given that fg(x) = 2g(f(x)), we start by finding each function:

Calculate fg(x):
- First calculate f(g(x)) = f(x² + 4) = 3(x² + 4) - 1 = 3x² + 12 - 1 = 3x² + 11.
- Next, substitute g(f(x)) = g(3x - 1) = (3x - 1)² + 4 = 9x² - 6x + 1 + 4 = 9x² - 6x + 5.
- Therefore, fg(x) = 2g(f(x)) becomes:
  $3x² + 11 = 2(9x² - 6x + 5)$ .
Expanding the right-hand side gives:
$3x² + 11 = 18x² - 12x + 10$ .
Rearrange to form a quadratic equation:
$0 = 18x² - 12x + 10 - 3x² - 11$ $0 = 15x² - 12x - 1$ .

This proves the required equation.

The functions f and g are such that f(x) = 3x - 1 and g(x) = x² + 4 (a) Find f⁻¹(x) Given that fg(x) = 2g(f(x)), (b) show that 15x² - 12x - 1 = 0. - Edexcel - GCSE Maths - Question 22 - 2019 - Paper 1

Worked Solution & Example Answer:The functions f and g are such that f(x) = 3x - 1 and g(x) = x² + 4 (a) Find f⁻¹(x) Given that fg(x) = 2g(f(x)), (b) show that 15x² - 12x - 1 = 0. - Edexcel - GCSE Maths - Question 22 - 2019 - Paper 1

Find f⁻¹(x)

show that 15x² - 12x - 1 = 0

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