How do you find the inverse of #y=cos x +3 # and is it a function?

Question

How do you find the inverse of  #y=cos x +3 # and is it a function?

Christian Antunez · Accepted Answer

The inverse function is piecewise and is #x= 2kpi + cos^(-1)(y-3), 2kpi <= x <= 2(k+1)pi, k = 0, +-1, +-2, +-3, ...#. See detailed explanation.

If y = f(x), the inverse is the explicit relation #x = f^(-1)(y)# = a function of y and, graphically, both give the same graph, in the same frame. Here, the inverse is #x = cos^(-1)( y - 3), 0 <= x <= pi#. The domain limits #[ 0 - pi ]# for x is attributed to the limits imposed in the conventional definition of inverse cosine. The graph for both y = 3 + cos x and its inverse #x = cos^(-1)( y - 3), 0 <= x <= pi # See the combined graph, for k = 0. graph{(y- 3 - cos x)(x-arccos (y-2.99))=0[0 3.14 1.9 4]} It is wrong to swap (x, y) as (y, x) and write the inverse as #y = cos^(-1)( x - 3 )#. Here, the two graphs are generally different, sans particular cases like y = 1 / x. See the combined graph for the wrong inverse. graph{(y- 3 - cos x)(y-arccos (x-3))=0[0 3.14 1.9 4]} For bijectivity ( one for one, either way), I have given piecewise definition for the inverse. See the graph below that is same for both. graph{(y- 3 - cos x)(x-arccos (y-3.01))=0[0 40 1.9 4.1] }

Greyson Brown · Answer

undefined To find the inverse of $ y = \cos(x) + 3 $, first, swap the roles of x and y. Then, solve for y.

$$
x = \cos(y) + 3
$$

Now, isolate $ \cos(y) $.

$$
\cos(y) = x - 3
$$

To find $ y $, take the inverse cosine of both sides.

$$
y = \arccos(x - 3)
$$

Whether this inverse is a function depends on the domain and range of the original function. Since the range of $ \cos(x) $ is from -1 to 1, when you add 3 to it, the range of $ y = \cos(x) + 3 $ is from 2 to 4. The domain of the inverse function will be the same as the range of the original function, and the range of the inverse function will be the same as the domain of the original function. Therefore, yes, the inverse is a function.