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)
To determine whether the function f(x) is even or odd, we need to analyze its properties for symmetry.
A) To check for even symmetry, we will substitute -x for x and compare f(x) and f(-x).
f(-x) = 5^((2(-x)^2)-1)^(1/2) (substituting -x for x)
= 5^(2x^2 - 1)^(1/2) (simplifying)
Now, let's compare f(x) and f(-x):
f(x) = 5^((2x^2)-1)^(1/2)
f(-x) = 5^(2x^2 - 1)^(1/2)
Since f(x) = f(-x), we can conclude that f(x) is an even function.
B) To find the domain of f, we need to identify any restrictions on the input values. In this case, there are no restrictions since the function involves only exponentiation and square roots, which can be evaluated for any real number. Therefore, the domain of f is all real numbers, or (-∞, ∞).
C) To determine the range of f, we need to find the set of all possible output values. In this case, because f(x) involves exponentiation and square roots, the output values will always be non-negative. Additionally, the base of the exponent is 5, which means the output values can be any non-negative real number. Therefore, the range of f is [0, ∞) or the set of all non-negative real numbers.
D) To find f'(x), we need to take the derivative of f(x) with respect to x. The given function is:
f(x) = 5^((2x^2)-1)^(1/2)
Using the chain rule, let's find the derivative step by step:
f'(x) = (1/2) * (5^((2x^2)-1))^(-1/2) * (2x) * (d/dx)((2x^2)-1)
= x * (5^((2x^2)-1))^(-1/2) * (4x)
Simplifying further, we have:
f'(x) = 4x^2 * (5^((2x^2)-1))^(-1/2)
Therefore, the derivative of f(x) with respect to x is f'(x) = 4x^2 * (5^((2x^2)-1))^(-1/2).