Find the range of f(x) for f(x) = e^3x + 1 for x element of real
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).
To find the range of f(x) = e^(3x) + 1, we need to determine the set of all possible values that f(x) can take.
First, notice that the function e^(3x) is always positive since e^t is positive for any real number t. Therefore, e^(3x) + 1 will always be positive as well.
To find the minimum value of f(x), we need to find the smallest possible value of e^(3x). Since e^(3x) is always positive, the smallest possible value occurs when e^(3x) = 0. However, e^(3x) can never be equal to 0 for any real value of x. Therefore, there is no minimum value for f(x), and the range of f(x) is greater than or equal to 1.
To find the maximum value of f(x), we can take the limit of f(x) as x approaches positive or negative infinity.
As x approaches positive infinity, e^(3x) will approach infinity as well. Therefore, the maximum value of f(x) is infinity.
As x approaches negative infinity, e^(3x) will approach 0 (since the exponent is negative), and adding 1 will give a minimum value of 1. Therefore, the minimum value of f(x) is 1.
Thus, the range of f(x) = e^(3x) + 1 for x element of real is [1, infinity).