range of f(x)= 5^(2x^2-1)^(1/2)
i know that the domain of the inverse would be the range of this and i've found the inverse but it has natural logs in it and i don't know how to find the domain with that.. i feel like i'm missing something simple
never mind i figured it out
To find the range of the function f(x) = 5^(2x^2 - 1)^(1/2), we need to consider the restrictions on the exponent and the base.
First, notice that the base of 5 is always positive. Therefore, the sign of our function will only depend on the exponent (2x^2 - 1)^(1/2).
Since the square root function (1/2 power) is always non-negative, the value of the exponent (2x^2 - 1) must be greater than or equal to zero:
2x^2 - 1 >= 0
Adding 1 to both sides:
2x^2 >= 1
Dividing both sides by 2:
x^2 >= 1/2
Taking the square root of both sides:
x >= ±√(1/2)
Therefore, the domain of the inverse function is x >= ±√(1/2).
Now, to find the range, we need to consider the value of 5^(2x^2 - 1)^(1/2) for different values of x.
When x = 0, the expression becomes 5^(2(0)^2 - 1)^(1/2) = 5^(-1)^(1/2) = 1/√5. So, 1/√5 is a possible output.
As x becomes more positive or negative, the exponent (2x^2 - 1) becomes larger, and thus, 5^(2x^2 - 1) increases. However, since we are taking the square root at the end, the range will be limited to non-negative values.
Therefore, the range of the function is [0, +∞), meaning all values greater than or equal to zero.
Note: To simplify your calculations, you can rewrite the expression as f(x) = √5^(2x^2 - 1). This way, the restriction on the domain becomes even clearer.