yeah thats the graph shape i get, its just i need to plot the other equation on the same graph, so i can find the place where they cross over, and they dont cross anywhere. the other equation is y = e^x and the line ive got already for that looks like its right, so where am i going wrong, or what am i missing ?
The problem did not say "use a graphing utility to produce the plot".
Whenever you want to sketch a curve you should start out by finding its x- and y-intercepts by setting y and x equal to 0 respectively. For you can't find any finite value, so you know the graph doesn't cut the x & y intercept.
Then you should determine its asymptotes, You can calculate the vertical asymptote by inspecting where the graph would be undefined (i.e., ), obviously in this case the denominator is zero at so you have a vertical asymptote there. In order to find the horizontal asymptote you need to compute the limits of your function as , if either limit has a finite value L, then the line y=L is a horizontal asymptote. , so your horizontal asymptote is at .
Using the values of y' you can then check where the graph is increasing (+ve) or decreasing (-ve). Values of y'' tells you where the graph is concave up or concave down. You have enough information to plot the graph.
As far as is concerned, you know that is the inverse of , so its graph can be obtained from that of by reflection in the line . Therefore its y-int is y=1 (as x=1 was the x-int of ln x), and the range of will be the set of all positive real numbers.