Let f be the function be defined by f(x)=5^((2x^2)-1)^(1/2)

A) is f an even or odd function? Justify answer.
B) find the domain of f.
C) find the range of f.
D) find f'(x)

A) To determine if a function is even or odd, we need to check if the function satisfies the properties of even or odd functions.

Even function:
A function f(x) is even if it satisfies the property f(x) = f(-x) for all x in the domain.
To check if f(x) is even, we substitute -x for x in the function and see if it remains unchanged.

f(-x) = 5^((2(-x)^2)-1)^(1/2)
= 5^((2x^2)-1)^(1/2)

Since f(x) = f(-x), the function remains unchanged when we substitute -x for x. Therefore, f(x) is an even function.

B) To find the domain of the function f(x), we need to determine the values of x for which the function is defined. In this case, since f(x) involves taking the square root of a value raised to a power, we need to ensure that the value inside the square root is non-negative.

The expression (2x^2) - 1 should be non-negative for the function to be defined.

2x^2 - 1 ≥ 0
2x^2 ≥ 1
x^2 ≥ 1/2

Taking the square root of both sides, we get:
|x| ≥ √(1/2)

So, the domain of f(x) is all real numbers where |x| is greater than or equal to the square root of 1/2.

C) To find the range of f(x), we need to determine the set of all possible output values of the function. In this case, we have a function involving the power of 5 and square roots.

Since the base of the exponential function f(x) is 5, and any number raised to a positive power is always positive, f(x) will always be positive for all x in its domain.

Therefore, the range of f(x) is all positive real numbers.

D) To find f'(x), the derivative of f(x), we need to use the power rule and chain rule.

f(x) = 5^((2x^2)-1)^(1/2)

Taking the derivative, we get:
f'(x) = (1/2) * (2x^2 - 1)^(-1/2) * (4x)
= 2x * (2x^2 - 1)^(-1/2)

Thus, f'(x) = 2x * (2x^2 - 1)^(-1/2).