The function f is defined by $$f: x ightarrow |2x - 5|, \, x \in \mathbb{R}$$ (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 6 - 2010 - Paper 5

To solve the equation $|2x - 5| = 15 + x$, we consider two cases based on the definition of absolute value:

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

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

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

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

3. Solutions:

   • The solutions to the equation are $x = 20$ and $x = -\frac{10}{3}$.

First compute $g(2)$ using $g(x) = x^2 - 4x + 1$:

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

2. Now compute fg(2) = f(g(2)) = f(-3): Using the function $f(x)$, $f(-3) = |2(-3) - 5| = |-6 - 5| = | -11 | = 11.$

3. Final Answer: Therefore, $fg(2) = 11$.

The function $g(x) = x^2 - 4x + 1$ is a quadratic function. To determine its range, we can complete the square:

1. Complete the Square: $g(x) = (x^2 - 4x + 4) - 4 + 1 = (x - 2)^2 - 3.$

2. Identify the Vertex: The vertex of the function is at $x = 2$, and therefore the minimum value can be found by substituting: $g(2) = -3.$

3. As x approaches the boundaries 0 and 5:

   • At $x = 0: g(0) = 1$
   • At $x = 5: g(5) = -4.$

4. Range of g: Since the function opens upwards, the range of $g$ on the interval (0, 5) is from the minimum value at the vertex to the maximum at the boundaries: $g: [-3, 1]$.