Given that
$f(x) = \ln x, \; x > 0$
sketch on separate axes the graphs of
(i) $y = f(x)$,
(ii) $y = |f(x)|$,
(iii) $y = -f(x) + 4$ - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 7
Question 2
Given that
$f(x) = \ln x, \; x > 0$
sketch on separate axes the graphs of
(i) $y = f(x)$,
(ii) $y = |f(x)|$,
(iii) $y = -f(x) + 4$.
Show, on each diagram, the ... show full transcript
Worked Solution & Example Answer:Given that
$f(x) = \ln x, \; x > 0$
sketch on separate axes the graphs of
(i) $y = f(x)$,
(ii) $y = |f(x)|$,
(iii) $y = -f(x) + 4$ - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 7
Step 1
(i) $y = f(x)$
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Answer
The function f(x)=lnx is defined for x>0.
Graph Behavior: The curve approaches the y-axis (vertical asymptote) as x→0. It crosses the x-axis at the point (1,0) since ln(1)=0. As x increases, f(x) increases without bound.
Asymptote: Vertical asymptote at x=0.
Step 2
(ii) $y = |f(x)|$
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Answer
Since f(x)=lnx is negative for 0<x<1, the absolute value graph reflects the part below the x-axis upwards.
Graph Behavior: This results in a continuous curve that is part of y=lnx for x>1 and part of a reflected line for 0<x<1. The graph crosses the x-axis at (1,0).
Asymptote: Vertical asymptote at x=0.
Step 3
(iii) $y = -f(x) + 4$
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Answer
Here, we examine the graph of y=−lnx+4.
Graph Behavior: This graph reflects and shifts the original logarithmic graph. It approaches the horizontal line y=4 as x increases, crossing the x-axis when −lnx+4=0, or at x=e4 approximately.
Asymptote: None for the x-axis, as this graph does not have a vertical asymptote but approaches y=4 as x→∞.
