Hi I have a question on graphing this function and its inverse
y =1/3 * e^(x+1)
In working this out I found that the inverse
X= ln(3y)-1
But I do not know how to draw both graph? Can someone help and this is due now L
Hi I have a question on graphing this function and its inverse
y =1/3 * e^(x+1)
In working this out I found that the inverse
X= ln(3y)-1
But I do not know how to draw both graph? Can someone help and this is due now L
The graph of 1/3 * e^(x+1) is obtained from the graph of e^x by shifting it to the left by 1 and squeezing it towards the horizontal axis.
The graph of ln(3x)-1 is obtained from the graph of ln(x) by squeezing it around the vertical axis and then lowering it by 1.
Note also that since these functions are inverse, their graph are symmetric with respect to the line y = x.
You can always use some software, such as WolframAlpha. One potential drawback WolframAlpha has is that it often shows graphs with different scales on the horizontal and vertical axes.