The function \(f(x)=\left|\left[\left(e^{2 x}-1\right) /\left(e^{2 x}+1\right)\right]\right|\) is (a) Increasing (b) Decreasing (c) Even (d) Strictly increasing

First, let's simplify the function \(f(x)\), since we'll need that for calculating the derivative. \(f(x) = \left|\left[\left(e^{2 x}-1\right) /\left(e^{2x}+1\right)\right]\right|\) Since the function is inside absolutes, we can write it as: \(f(x) = \left|\frac{e^{2x} - 1}{e^{2x} + 1}\right|\) Now, to determine if it's increasing or decreasing, we'll need to find its derivative with respect to \(x\). The derivative is: \(f'(x) = \frac{d}{dx} \left(\left|\frac{e^{2x} - 1}{e^{2x} + 1}\right|\right)\)

Since we have the absolute value of a function as our main function, we can use the chain rule to find its derivative. The chain rule states that if we have a function \(u(v)\) then its derivative with respect to \(x\) is given by: \(\frac{d}{dx} u(v) = u'(v) \cdot v'(x)\) So, for our function, we have: \(u(v) = |v|\) \(u'(v) = v/|v|\) only if v ≠ 0 and \(v = \frac{e^{2x} - 1}{e^{2x} + 1}\) \(v'(x) = \frac{2e^{2x}(e^{2x}+1) - 2e^{2x}(e^{2x}-1)}{(e^{2x}+1)^{2}}\) Now let's combine \(u'(v)\) and \(v'(x)\) to find \(f'(x)\) as follows: \(f'(x) = u'(v)\cdot v'(x) = \frac{v}{|v|} \cdot \frac{2e^{2x}(e^{2x}+1) - 2e^{2x}(e^{2x}-1)}{(e^{2x}+1)^{2}}\) Simplify: \(f'(x) = \frac{v}{|v|} \cdot \frac{4e^{2x}}{(e^{2x} + 1)^2}\)

Since we have our derivative, we can now evaluate the properties. (a) Increasing: We observe that the derivative \(f'(x)\) has a positive value, as both factors are positive for all \(x\) (the numerator is the product of positive numbers and the denominator is a square). Therefore, \(f'(x) > 0\) for all \(x\), which means the function is increasing. (b) Decreasing: We already determined that the function is increasing, so it cannot be decreasing. (c) Even: To check if the function is even, we need to evaluate if: \(f(x) = f(-x) ⟹ \left|\frac{e^{2x} - 1}{e^{2x} + 1}\right| = \left|\frac{e^{-2x} - 1}{e^{-2x} + 1}\right|\) Simplifying the expression in the right side: \(\left|\frac{e^{-2x} - 1}{e^{-2x} + 1}\right| = \left|\frac{1 - e^{2x}}{1 + e^{2x}}\right| = \left|-\frac{e^{2x} - 1}{e^{2x} + 1}\right|\) So, both the expressions are equivalent since they only differ in sign, we can conclude that the function is even, as \(f(x) = f(-x)\). (d) Strictly increasing: Since \(f'(x)>0\) for all values of \(x\), the function is strictly increasing.

The given function \(f(x) = \left|\left[\left(e^{2 x}-1\right) /\left(e^{2x}+1\right)\right]\right|\) is (a) increasing, (c) even, and (d) strictly increasing. It is not (b) decreasing.