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Figure 1 shows a sketch of the curve with equation $y = f(x)$, $x > 0$, where $f$ is an increasing function of $x$ - Edexcel - A-Level Maths Pure - Question 4 - 2013 - Paper 8

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Figure-1-shows-a-sketch-of-the-curve-with-equation-$y-=-f(x)$,-$x->-0$,-where-$f$-is-an-increasing-function-of-$x$-Edexcel-A-Level Maths Pure-Question 4-2013-Paper 8.png

Figure 1 shows a sketch of the curve with equation $y = f(x)$, $x > 0$, where $f$ is an increasing function of $x$. The curve crosses the $x$-axis at the point $(1, ... show full transcript

Worked Solution & Example Answer:Figure 1 shows a sketch of the curve with equation $y = f(x)$, $x > 0$, where $f$ is an increasing function of $x$ - Edexcel - A-Level Maths Pure - Question 4 - 2013 - Paper 8

Step 1

Sketch the curve with equation y = f(2x), x > 0

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Answer

To sketch the curve for y=f(2x)y = f(2x), we need to take into account that the function is compressed horizontally by a factor of 2. The curve will cross the xx-axis at the point (0.5,0)(0.5, 0) since for x=0.5x=0.5, f(20.5)=f(1)=0f(2*0.5)=f(1)=0. The overall shape should be an increasing function starting from the asymptote at x=0x=0, gently rising and crossing the xx-axis at (0.5,0)(0.5, 0).

Step 2

Sketch the curve with equation y = |f(x)|, x > 0

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Answer

For the function y=f(x)y = |f(x)|, the result is that it remains non-negative. Since f(x)f(x) is positive for x>1x > 1, the graph will mimic the original curve for x>1x > 1. However, at x=1x=1, where f(x)f(x) crosses the xx-axis, the behavior changes: instead of going negative, the graph will turn up, creating a cusp at the point (1,0)(1, 0). Thus, the curve will approach (1,0)(1, 0) from above, reflecting the point rather than continuing below.

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