I think this is a easy trick question, so I probably know the answer but I just wanted to double check.
What is the domain and range of the function f(x) = 3^x
Find the f^-1 and its domain and range?
Thanks for your help.
domain: any real number
range: any real number > 0
for the domain of the inverse function, since you interchange the x and the y,
the range of the old becomes the domain of the inverse, and vice versa.
Thank you - that is exactly what I thought!
To find the domain and range of the function f(x) = 3^x, we need to understand the properties of an exponential function.
The domain of an exponential function is all real numbers, as there are no restrictions on the base (3 in this case) or the exponent (x). In other words, x can take any real value.
The range of an exponential function depends on the base. In the case of f(x) = 3^x, where the base is 3, the range is all positive real numbers greater than 0.
Now, to find the inverse of the function f(x), denoted as f^-1(x), we need to interchange the roles of x and y in the original function and solve for y.
Starting from f(x) = 3^x, we replace f(x) with y:
y = 3^x
Next, we interchange x and y:
x = 3^y
To solve for y, we can take the logarithm (base 3) of both sides of the equation:
log₃(x) = log₃(3^y)
Using the logarithmic property, we can bring down the exponent:
log₃(x) = y * log₃(3)
Simplifying further, as log₃(3) equals 1:
log₃(x) = y
So, the inverse function is f^-1(x) = log₃(x).
The domain of the inverse function is all positive real numbers greater than 0, as the logarithm of a negative number or zero is undefined.
The range of the inverse function is all real numbers, as the logarithm can produce any real number output for positive inputs.
Hope this helps! Let me know if you have any further questions.