You can have a quick answer by graphing the function.
Its domain is all real numbers.
Firstly graph the function F1(x) = 34(x-5)
The function has the x-axis as horizontal asymptote and as x tends to ∞ the y's tend to ∞.
Then graph the function F2(x) =− 34(x-5) by taking the image of F1 about the x-axis.
Then push the graph of F2 to units up and you get the graph of f(x) =2− 34(x-5).
By now you will have understood that the range is (−∞,2).
Another approach would be to notice that the function is one-to-one. Then find the inverse and hence the domain of the inverse is the range of the original.
x= 2− 34(y-5)
34(y−5) = 2−x
Taking the log3 of both sides
4(y−4) = log3 (2−x)
y= (1/4) [ 16 + log3 (2−x) ]
ƒ-1(x) = (1/4) [ 16 + log3 (2−x) ]
Now in order for the log3 (2−x) to be defined 2−x >0 . The domain of ƒ-1 and range of ƒ is (−∞,2 )
