Find the range of f(x) for f(x) = e^3x + 1 for x element of real

Answer 1

To find the range of f(x) = e^(3x) + 1, we need to determine the set of all possible output values.

Since e^3x is always positive for any real value of x, the smallest possible value for e^3x would be when x = -∞. In this case, e^3x approaches 0 as x approaches -∞. So, the smallest possible value for f(x) would be 1 when x = -∞.

As x increases, e^3x will also increase, resulting in f(x) increasing as well. There is no upper bound for e^3x, so f(x) will continue to increase indefinitely.

Therefore, the range of f(x) = e^3x + 1 is (1, ∞), where ( denotes an exclusive boundary, meaning 1 and ∞ are not included in the range).