2.1 just plug and chug
the inverse is, of course, x = (1/2)^y
That is, y = log1/2(x) = -logx/log2
see what you can do with the rest
Refer to the sketch given below of y=g(x) and y=g^-1(x), where g(x)=b^x
T is the point of intersection of 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).
1 answer