The domain of the inverse function, f^-1, is the range of the original function, f(x).

For the function f(x) = √(x + 5) + 2, the smallest value that the function can take is when x = -5, because the square root of a negative number is not a real number. When x = -5, f(x) = √(0) + 2 = 2.

Since the square root function and addition by 2 both increase as x increases, the function f(x) will continue to increase as x increases. Therefore, the range of f(x) is all real numbers greater than or equal to 2.

So, the domain of f^-1 is all real numbers greater than or equal to 2.

Question

The domain of the inverse function, f^-1, is the range of the original function, f(x).

For the function f(x) = √(x + 5) + 2, the smallest value that the function can take is when x = -5, because the square root of a negative number is not a real number. When x = -5, f(x) = √(0) + 2 = 2.

Since the square root function and addition by 2 both increase as x increases, the function f(x) will continue to increase as x increases. Therefore, the range of f(x) is all real numbers greater than or equal to 2.

So, the domain of f^-1 is all real numbers greater than or equal to 2.

Knowee AI · Accepted Answer

The domain of the inverse function, f^-1, is the range of the original function, f(x).

For the function f(x) = √(x + 5) + 2, the smallest value that the function can take is when x = -5, because the square root of a negative number is not a real number. When x = -5, f(x) = √(0) + 2 = 2.

Since the square root function and addition by 2 both increase as x increases, the function f(x) will continue to increase as x increases. Therefore, the range of f(x) is all real numbers greater than or equal to 2.

So, the domain of f^-1 is all real numbers greater than or equal to 2.

For f(x) = √(x + 5) + 2, what is the domain of f-1

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