The function f is defined by $$f : x \mapsto |2x - 5|, \; x \in \mathbb{R}$$ (a) Sketch the graph with equation $y = f(x)$, showing the coordinates of the points where the graph cuts or meets the axes - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 5

We need to solve the equation: $|2x - 5| = 15 + x$

This absolute value equation can be split into two cases:

1. Case 1: $2x - 5 = 15 + x$

   • Rearranging gives: $2x - x = 15 + 5\implies x = 20$

2. Case 2: $2x - 5 = -(15 + x)$

   • Rearranging gives: $2x - 5 = -15 - x \implies 2x + x = -15 + 5 \implies 3x = -10 \implies x = -\frac{10}{3}$

Thus, the solutions are: $x = 20 \text{ and } x = -\frac{10}{3}.$

To find $fg(2)$, we first calculate $g(2)$ using the function: $g(x) = x^2 - 4x + 1$

1. Substitute $x = 2$: $g(2) = 2^2 - 4(2) + 1 = 4 - 8 + 1 = -3$

2. Now we find $f(g(2)) = f(-3)$: $f(-3) = |2(-3) - 5| = |-6 - 5| = |-11| = 11$

Therefore, $fg(2) = 11.$

To determine the range of the function $g(x) = x^2 - 4x + 1$, we can complete the square:

1. Rewrite $g(x)$: $g(x) = (x^2 - 4x + 4 - 4 + 1) = (x - 2)^2 - 3$

2. The vertex form indicates that the minimum value occurs at $x = 2$, giving: $g(2) = -3$

3. Since $g(x)$ is a quadratic function that opens upwards, the range is: $[-3, \infty)$ over the restricted domain $0 \leq x \leq 5$ could also be checked by evaluating endpoints: $g(0) = 1 \quad \text{and} \quad g(5) = 16$ Thus, considering the range of endpoints: $[-3, 16]$.