Let's analyze the function $f(x) = \frac{1}{(x^4 + 1)^2}$ step-by-step to answer each of the given parts.

Step 1: Find the First Derivative

To find the critical points and determine where the function is increasing or decreasing, we first need to calculate the first derivative, $f'(x)$.

Given: $f(x) = (x^4 + 1)^{-2}$

Apply the chain rule: $f'(x) = -2(x^4 + 1)^{-3} \cdot (4x^3)$ $f'(x) = -\frac{8x^3}{(x^4 + 1)^3}$

Step 2: Find the Second Derivative

To identify inflection points and determine concavity, we need the second derivative, $f''(x)$.

First, rewrite the first derivative: $f'(x) = -\frac{8x^3}{(x^4 + 1)^3}$

Now, apply the quotient rule to find $f''(x)$: $f''(x) = \frac{d}{dx} \left( -\frac{8x^3}{(x^4 + 1)^3} \right)$

Quotient rule: $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$, where $u = -8x^3$ and $v = (x^4 + 1)^3$.

Compute $u'$ and $v'$: $u' = -24x^2, \quad v' = 3(x^4 + 1)^2 \cdot 4x^3 = 12x^3 (x^4 + 1)^2$

Substitute into the quotient rule: $f''(x) = \frac{(-24x^2)(x^4 + 1)^3 - (-8x^3)(12x^3(x^4 + 1)^2)}{(x^4 + 1)^6}$

Simplify the numerator: $f''(x) = \frac{-24x^2(x^4 + 1)^3 + 96x^6(x^4 + 1)^2}{(x^4 + 1)^6}$ $f''(x) = \frac{-24x^2(x^4 + 1)^2 \left( x^4 + 1 - 4x^4 \right)}{(x^4 + 1)^6}$ $f''(x) = \frac{-24x^2(x^4 + 1)^2 (-3x^4 + 1)}{(x^4 + 1)^6}$

Simplify further: $f''(x) = \frac{72x^2(x^4 + 1)^2(x^4 - \frac{1}{3})}{(x^4 + 1)^6}$ $f''(x) = \frac{72x^2(x^4 - 1/3}{(x^4 + 1)^4}$

The Sign of f´´ is helpful. Now and return find derive I report analysis.