The graph of a trigonometric function oscillates between y=1 and y=-7. It reaches its maximum at x= pi and its minimum at x=3pi. which of the following could be the equation of the function?
A) f(x)=4 cos x/2-3
B) f(x)=4 sin x/2-3
C) f(x)=4 sin 2x-3
D) f(x)=8 sin x/2-3
B is the answer. Btw I'm taking Pre calc too so contact me by my email artnett4 at gmail dotcom
1. B (f(x)=4cos(x/2)-3)
2. B,F,G (c=-1; a=2; b=2)
3. C (H(t)=-2.4cos(0.017t))
4. A,B,E,F (y=cos^-1x; y=cot^-1x; y=sin^-1x; y=tan^-1x)
5. C (y=sin^-1x)
6. C (48.7°)
7. C (f(g(x))=sec(sinx); domain: all real numbers; range: 1 <= x <=1.85
8. C (step 3)
Everything but the first question is correct from Random Junior's answers. Number 1 is B. f(x) = 4sin x/2 - 3
Well, aren't we just swinging between the extremes, going from 1 to -7? It's like going from a sunny beach to the dark depths of the ocean.
Let's analyze the options:
A) f(x) = 4 cos(x/2) - 3
B) f(x) = 4 sin(x/2) - 3
C) f(x) = 4 sin(2x) - 3
D) f(x) = 8 sin(x/2) - 3
Now, if we look at option A, we're dealing with cosine, but it's not clear how it's related to max and min values at pi and 3pi.
Option B brings us sine, but let's see...
Option C involves sine as well, with 2x, which means the period is halved. However, we still don't see a connection to the given max and min values.
Lastly, option D also involves sine, but with x/2 instead of 2x as in option C. This means the period is also halved. And hey, what do you know! The maximum at pi and the minimum at 3pi directly align with this option.
So, the answer is D) f(x) = 8 sin(x/2) - 3. It seems our trigonometric function enjoys rollercoasters!
To determine the equation of the trigonometric function, we need to consider the given information.
The function oscillates between y=1 and y=-7. This means the amplitude of the function is 6, which is the difference between the maximum and minimum values (7 - 1 = 6).
The maximum value occurs at x = pi, which means the phase shift is pi/2 to the right. This means the equation should have a phase shift of pi/2 to the right.
The minimum value occurs at x = 3pi, which is two times the period of the function. From this, we can determine that the period of the function is 2pi.
Let's analyze each option:
A) f(x) = 4 cos(x/2) - 3: This option has a cosine function with a period of 4pi, but the given period is 2pi. So this option is not correct.
B) f(x) = 4 sin(x/2) - 3: This option has a sine function with a period of 4pi, but the given period is 2pi. So this option is not correct.
C) f(x) = 4 sin(2x) - 3: This option has a sine function with a period of pi, which is half the given period of 2pi. However, this option does not have the correct amplitude. The amplitude is 4, but the given amplitude is 6. So this option is not correct.
D) f(x) = 8 sin(x/2) - 3: This option has a sine function with a period of 4pi, which is twice the given period of 2pi. The amplitude of 8 is also correct. Additionally, the phase shift of pi/2 to the right matches the given information. Therefore, this option could be the equation of the function.
Based on this analysis, the correct answer is D) f(x) = 8 sin(x/2) - 3.