Require assistance with algebraic functions.

Question

y=g(x) and y=g^-1(x), where g(x)=b^x

T is the point of intersection for both graphs.

The point K(-1;2) lies on the graph of g.

2.1 Show that b=1/2

2.2 Determine the equation of g^-1(x) in the form of y=...

2.3 The x-coordinate of T is 0.62. Give the y-coordinate of T

2.4 For which values of x will g^-1(x)>0

2.5 h(x) is the reflection of g(x) about the x-axis. Write down the equation of h(x).

Answer 1

g(x) = b^x = y

plug in the point (-1,2) x =-1, y =2

b^-1 = 2

1/b = 2

b =1/2

g(x) = y = b^x = .5^x

switch x and y and solve for the new y = g^-1(x)

x = b^y

take logs of both sides

log_bx = y = g^-1(x)

y = log_.5(x) is the inverse function

T is the intersection point?

where g(x) = g^-1(x)

where log_.5(x) = .5^x

if .62 is the x coordinate of T

plug .62 into (1/2)^x to get (.5)^.62 = about 0 .7 = the y coordinate of the "intersection pont"

or plug .62 into log_.5x = log_.5.62 = the y coordinate

= ln.62/ln.5 = -.478/-.7 =about 0.7

intersection point T (0.6, 0.7)

g^-1(x)>0 when x>1

h(x) is a reflection across the x axis of g(x) = b^x = .5^x

h(x) = g(-x) = .5^-x