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The function f is defined by $f: x \mapsto |2x-5|, \; x \in \mathbb{R}$ - Edexcel - A-Level Maths Pure - Question 5 - 2010 - Paper 5

Question 5

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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 t... show full transcript

Worked Solution & Example Answer:The function f is defined by $f: x \mapsto |2x-5|, \; x \in \mathbb{R}$ - Edexcel - A-Level Maths Pure - Question 5 - 2010 - Paper 5

Step 1

Sketch the graph with equation y=f(x), showing the coordinates of the points where the graph cuts or meets the axes.

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Answer

To sketch the graph of the function $f(x) = |2x - 5|$ , we first find the points where it intersects the axes.

Finding the x-intercepts: Set $f(x) = 0$ : [ |2x - 5| = 0 ] This leads to: [ 2x - 5 = 0 \Rightarrow x = 2.5 ]
Finding the y-intercept: Set $x = 0$ : [ f(0) = |2(0) - 5| = 5 ] Therefore, the y-intercept is at (0, 5).
Behaviors of function: The graph is V-shaped with the vertex at $(2.5, 0)$ . For $x < 2.5$ , $f(x) = 5 - 2x$ and for $x > 2.5$ , $f(x) = 2x - 5$ . The corners are thus (0, 5), (2.5, 0).

To sketch:

Plot the points (0, 5) and (2.5, 0).
Join these points, reflecting the absolute value's V shape.

Step 2

Solve f(x)=15+x.

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Answer

We need to find the value of $x$ for which:

$|2x - 5| = 15 + x.$

We solve this by considering two cases based on the definition of absolute value:

Case 1: $2x - 5 = 15 + x$ [ 2x - x = 15 + 5\Rightarrow x = 20 ]
Case 2: $2x - 5 = -(15 + x)$ [ 2x - 5 = -15 - x \Rightarrow 3x = -10 \Rightarrow x = -\frac{10}{3} ]

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

Step 3

Find fg(2).

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Answer

We need to calculate $fg(2)$ using the functions defined:

First, find g(2): [ g(x) = x^2 - 4x + 1
g(2) = (2)^2 - 4(2) + 1 = 4 - 8 + 1 = -3 ]
Next, find f(g(2)) = f(-3): [ f(-3) = |2(-3) - 5| = |-6 - 5| = | -11 | = 11 ]

Thus, $fg(2) = 11$ .

Step 4

Find the range of g.

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Answer

To find the range of the function $g(x) = x^2 - 4x + 1$ for $0 \leq x \leq 5$ , we can use the vertex form of a quadratic.

Complete the square: [ g(x) = (x^2 - 4x + 4) - 4 + 1 = (x - 2)^2 - 3 ]
Find the vertex: The vertex occurs at $x = 2$ , giving: [ g(2) = (2 - 2)^2 - 3 = -3. ]
Evaluate endpoints: At $x=0$ : [ g(0) = 0^2 - 4(0) + 1 = 1 ] At $x=5$ : [ g(5) = 5^2 - 4(5) + 1 = 1. ]

Therefore, the minimum value is $-3$ and the maximum is $1$ . So, the range of $g$ is: $[-3, 1].$

The function f is defined by $f: x \mapsto |2x-5|, \; x \in \mathbb{R}$ - Edexcel - A-Level Maths Pure - Question 5 - 2010 - Paper 5

Worked Solution & Example Answer:The function f is defined by $f: x \mapsto |2x-5|, \; x \in \mathbb{R}$ - Edexcel - A-Level Maths Pure - Question 5 - 2010 - Paper 5

Sketch the graph with equation y=f(x), showing the coordinates of the points where the graph cuts or meets the axes.

Solve f(x)=15+x.

Find fg(2).

Find the range of g.

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