which pair of functions is not a pair of inverse functions?

a. f(x)= x+1/6 and g(x)= 6x-1

b. f(x)= x-4/19 and g(x)= 19x+4

c. f(x)= x5 and g(x)= 5√x

d. f(x)= x/x + 20 and g(x)= 20x/x-1
you!

Option in D is not an inverse function.

Step-by-step explanation:

A. Say,

⇒ 6y = x + 1

⇒ x = 6y - 1

Here, g(x) = 6x - 1, if g(x) is the inverse function of f(x).

B. Let us assume,

⇒ 19y = x - 4

⇒ x = 19x + 4

If g(x) is the inverse function of f(x), then, g(x) = 19x + 4

C. Let

⇒

Here, , if g(x) is the inverse function of f(x).

D. Let

⇒ yx + 20y = x

⇒ x(y - 1) = - 20y

⇒

Now, if g(x) is the inverse function of f(x), then,

Therefore, option in D is not an inverse function. (Answer)

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.