Question #a818a

Here is the graph:

This looks like a cosine function that has been multiplied by an amplitude and phase shift to the right:

#f(x) = Acos(x - phi) = sin(x) + cos(x)#

My graphing tool allows me to obtain values of points and I can tell you that #A = sqrt(2)#

#cos(x - phi) = 1/sqrt(2)sin(x) + 1/sqrt(2)cos(x)#

This fits the trigonometric identity:

#cos(x - phi) = sin(x)sin(phi) + cos(x)cos(phi)#

where #cos(phi) = sin(phi) = 1/sqrt(2)#

This happens at #phi = pi/4#

The graphing tool confirms that the x coordinates are shift by #pi/4#.

#f(x) = sqrt(2)cos(x - pi/4)#

Substitute #f^-1(x)# for every x:

#f(f^-1(x)) = sqrt(2)cos(f^-1(x) - pi/4)#

The left side becomes x by definition:

#x = sqrt(2)cos(f^-1(x) - pi/4)#

Make the #sqrt(2)# on the right disappear by multiplying both sides by #sqrt(2)/2#:

#sqrt(2)/2x = cos(f^-1(x) - pi/4)#

Use the inverse cosine on both sides:

#cos^-1(sqrt(2)/2x) = f^-1(x) - pi/4#

Solve for #f^-1(x)#:

#f^-1(x) = cos^-1(sqrt(2)/2x) + pi/4#

To confirm that this is truly an inverse, verify that #f(f^-1(x)) = f^-1(f(x)) = x#