course mth 172 i put all my info on it this time.
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RESPONSE --> 1 = 8 cos(2x+1) - 3 4=8 cos(2x+1) 32=cos(2x+1) ??
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20:38:48 ** 2x + 1 = `pi/3, or `pi / 3 + 2 `pi, or `pi / 3 + 4 `pi, ... or `pi / 3 + 2n `pi, ... with n = an integer. So x = (`pi / 3 - 1) / 2. Or x = (`pi / 3 + 2 `pi - 1) / 2 = (`pi / 3 - 1) / 2 + `pi. Or x = (`pi / 3 + 4 `pi - 1) / 2 = (`pi / 3 - 1) / 2 + 2 `pi. … Or x = (`pi / 3 + 2 n `pi - 1) / 2 = (`pi / 3 - 1) / 2 + n `pi ... x = (`pi / 3 - 1) / 2 + n `pi, n an integer, is the general solution. Since 0 (`pi / 3 - 1) / 2 < `pi he first two solutions are between 0 and 2 `pi, and are the only solutions between 0 and 2 `pi. We generally want at least the solutions between 0 and 2 `pi. ** RESPONSE i still don't understand how you get 2x + 1 = `pi/3 out of 1 = 8 cos(2x+1) - 3 does'pi mean the dirivitive of pi? please explain in simple terms. i hate this computer learning bullshit, i need some damn pictures, and need to get this torture over with so i can take calculus 2 in a classroom next spring. i have put this off as long as i could, but i am ready to finish this.
20:40:25 problem 1.5.42 (was 1.9.34) arccos fndescribe the graph of the arccos or cos^-1 function, including a statement of its domain and range, increasing/decreasing behavior, asymptotes if any, and concavityexplain why the domain and range are as you describe......!!!!!!!!...................................
RESPONSE --> ok. arccos? fndescribe? i had trig about 7 years ago, so i need help
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20:41:48 ** To get the graph of arccos(x) we take the inverse of the graph of cos(x), restricting ourselves to 0 < = x < = `pi, where y = cos(x) takes each possible y value exactly once. The graph of this portion of y = cos(x) starts at (0, 1), decreases at an increasing rate (i.e., concave down) to (`pi/2, 0) the decreases at a decreasing rate (i.e., concave up) to (`pi, -1). The inverse function reverses the x and y values. Since y values of the cosine function range from -1 to 1, the x values of the inverse function will run from -1 to 1. The graph of the inverse function therefore runs from x = -1 to x = 1. The graph of the inverse function starts at (-1, `pi) and runs to (0, `pi/2), reversing the values of y = cos(x) between (`pi/2, 0) and (`pi, -1). Since the graph of y = cos(x) is decreasing at a decreasing rate (i.e., concave up) between these points, the graph of the inverse function will be decreasing at a decreasing rate (i.e., concave up) over the corresponding region. The graph of the inverse function continues from (0, `pi/2) to (1, 0), where it decreases at an increasing rate. The values of the inverse cosine function range from 0 to `pi; the domain of this function is [-1, 1].**
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RESPONSE --> help? To start with think of a table for the cosine function from x = 0 to x = +pi. We use these values because the value of the cosine goes from 1 to -1, i.e., through its full range, between these x values.
x cos(x) 0 1 pi/4 .71 pi/2 0 3pi/4 -.71 pi -1. If we reverse the columns of this table we get a partial table for the inverse cosine function. To avoid confusion with x in the above table we'll use z for the variable: z arcCos(z) 1 0 .71 pi/4 0 pi/2 -.71 3pi/4 -1 pi Graph both the cos(x) vs. x and arcCos(z) vs. z functions according to these tables. Then make an extended table for cos(x) using x = 0, pi/6, pi/3, pi/2, 2pi/3, 5pi/6 and pi. Finally reverse this table to get a table for arcCos(z) vs. z. These tables should allow you to refine your graphs. Finally follow the same scheme for sin(x) vs. x for x = -pi/2 to pi/2, reversing columns to get arcsin(z) vs. z. Send me a description of what you get and I'll tell you more so arc means inverse of. now what about findscribe?"
course mth 172 i put all my info on it this time.
......!!!!!!!!...................................
RESPONSE --> 1 = 8 cos(2x+1) - 3 4=8 cos(2x+1) 32=cos(2x+1) ??
.................................................
......!!!!!!!!...................................
20:38:48 ** 2x + 1 = `pi/3, or `pi / 3 + 2 `pi, or `pi / 3 + 4 `pi, ... or `pi / 3 + 2n `pi, ... with n = an integer. So x = (`pi / 3 - 1) / 2. Or x = (`pi / 3 + 2 `pi - 1) / 2 = (`pi / 3 - 1) / 2 + `pi. Or x = (`pi / 3 + 4 `pi - 1) / 2 = (`pi / 3 - 1) / 2 + 2 `pi. … Or x = (`pi / 3 + 2 n `pi - 1) / 2 = (`pi / 3 - 1) / 2 + n `pi ... x = (`pi / 3 - 1) / 2 + n `pi, n an integer, is the general solution. Since 0 (`pi / 3 - 1) / 2 < `pi he first two solutions are between 0 and 2 `pi, and are the only solutions between 0 and 2 `pi. We generally want at least the solutions between 0 and 2 `pi. ** RESPONSE i still don't understand how you get 2x + 1 = `pi/3 out of 1 = 8 cos(2x+1) - 3 does'pi mean the dirivitive of pi? please explain in simple terms. i hate this computer learning bullshit, i need some damn pictures, and need to get this torture over with so i can take calculus 2 in a classroom next spring. i have put this off as long as i could, but i am ready to finish this.
20:40:25 problem 1.5.42 (was 1.9.34) arccos fndescribe the graph of the arccos or cos^-1 function, including a statement of its domain and range, increasing/decreasing behavior, asymptotes if any, and concavityexplain why the domain and range are as you describe......!!!!!!!!...................................
RESPONSE --> ok. arccos? fndescribe? i had trig about 7 years ago, so i need help
.................................................
......!!!!!!!!...................................
20:41:48 ** To get the graph of arccos(x) we take the inverse of the graph of cos(x), restricting ourselves to 0 < = x < = `pi, where y = cos(x) takes each possible y value exactly once. The graph of this portion of y = cos(x) starts at (0, 1), decreases at an increasing rate (i.e., concave down) to (`pi/2, 0) the decreases at a decreasing rate (i.e., concave up) to (`pi, -1). The inverse function reverses the x and y values. Since y values of the cosine function range from -1 to 1, the x values of the inverse function will run from -1 to 1. The graph of the inverse function therefore runs from x = -1 to x = 1. The graph of the inverse function starts at (-1, `pi) and runs to (0, `pi/2), reversing the values of y = cos(x) between (`pi/2, 0) and (`pi, -1). Since the graph of y = cos(x) is decreasing at a decreasing rate (i.e., concave up) between these points, the graph of the inverse function will be decreasing at a decreasing rate (i.e., concave up) over the corresponding region. The graph of the inverse function continues from (0, `pi/2) to (1, 0), where it decreases at an increasing rate. The values of the inverse cosine function range from 0 to `pi; the domain of this function is [-1, 1].**
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RESPONSE --> help? To start with think of a table for the cosine function from x = 0 to x = +pi. We use these values because the value of the cosine goes from 1 to -1, i.e., through its full range, between these x values.
x cos(x) 0 1 pi/4 .71 pi/2 0 3pi/4 -.71 pi -1. If we reverse the columns of this table we get a partial table for the inverse cosine function. To avoid confusion with x in the above table we'll use z for the variable: z arcCos(z) 1 0 .71 pi/4 0 pi/2 -.71 3pi/4 -1 pi Graph both the cos(x) vs. x and arcCos(z) vs. z functions according to these tables. Then make an extended table for cos(x) using x = 0, pi/6, pi/3, pi/2, 2pi/3, 5pi/6 and pi. Finally reverse this table to get a table for arcCos(z) vs. z. These tables should allow you to refine your graphs. Finally follow the same scheme for sin(x) vs. x for x = -pi/2 to pi/2, reversing columns to get arcsin(z) vs. z. Send me a description of what you get and I'll tell you more so arc means inverse of. now what about findscribe?"