i don't know if its just undefined or if it's possible please explain
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Why would y = arccos(x/3) be undefined?
Anyways, if you solve y = arccos(x/3) for x, you get:
cos(y) = x/3, by taking the cosine of both sides
==> x = 3cos(y).
However, the range of arccos(x/3) is [0, π], so y can only be on [0, π].
If you can graph the cosine function, just graph x = 3cos(y) on the interval [0, π]; the resulting graph will be the graph for y = arccos(x/3).
I hope this helps!
Anyways, if you solve y = arccos(x/3) for x, you get:
cos(y) = x/3, by taking the cosine of both sides
==> x = 3cos(y).
However, the range of arccos(x/3) is [0, π], so y can only be on [0, π].
If you can graph the cosine function, just graph x = 3cos(y) on the interval [0, π]; the resulting graph will be the graph for y = arccos(x/3).
I hope this helps!
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To get the equation of an inverse of a function, plug x in for y and y in for x and solve for y
x = arccos(y/3) To get rid of arccos, we need to take cos
cos (x) = cos (arccos (y/3)) This cancels out the arccos so what's left is
cos (x) = y/3
y = 3 cos (x)
x = arccos(y/3) To get rid of arccos, we need to take cos
cos (x) = cos (arccos (y/3)) This cancels out the arccos so what's left is
cos (x) = y/3
y = 3 cos (x)