Sketch each of the following graphs[ u can don't sketch, but describe clearly about the amplitude, period, min and max point clearly] for 0
Please show the steps that to find the amplitude and so on so that I can comprehend. THANKS
A)y=2/(0.5 + |2cosx|)
B)y=1/(1-0.5|sin x|)
* INVOLVES MODULUS FUNCTIONS
Please show the steps that to find the amplitude and so on so that I can comprehend. THANKS
A)y=2/(0.5 + |2cosx|)
B)y=1/(1-0.5|sin x|)
* INVOLVES MODULUS FUNCTIONS
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(A) It's a periodic function, and its period is pi,
but I wouldn't necessarily say it has an "amplitude," because it's not shaped like a sine wave.
It has sharp points (upward) at x=pi/2 and x=3pi/2, and these are the maxima,
where obviously y = 2/0.5 = 4. The minima occur at x = 0,pi,2pi, where y = 2/(0.5+2) = 0.8.
These are not sharp points, they are sort of shallow valleys.
The range is [0.8,4].
(B) For this one, you have sharp points downward at x = 0,pi,2pi, where y = 1,
and these are the minima. You have round-topped maxima at x = pi/2 and 3pi/2,
where y = 1 / (1/2) = 2. The period is pi and the range is [1,2].
The reason for the sharp corners in both of these graphs is the sudden change of direction provided by the absolute value sign.
but I wouldn't necessarily say it has an "amplitude," because it's not shaped like a sine wave.
It has sharp points (upward) at x=pi/2 and x=3pi/2, and these are the maxima,
where obviously y = 2/0.5 = 4. The minima occur at x = 0,pi,2pi, where y = 2/(0.5+2) = 0.8.
These are not sharp points, they are sort of shallow valleys.
The range is [0.8,4].
(B) For this one, you have sharp points downward at x = 0,pi,2pi, where y = 1,
and these are the minima. You have round-topped maxima at x = pi/2 and 3pi/2,
where y = 1 / (1/2) = 2. The period is pi and the range is [1,2].
The reason for the sharp corners in both of these graphs is the sudden change of direction provided by the absolute value sign.