Consider the function  f(x)=1(x4+1)2
  .

(a) Give the exact  x
 value of a non-stationary inflection point (an inflection point which is not a critical point)
Preview  

(b) Give an  x
 value where the function is increasing, but concave down.
Preview  

(c) Give an  x
 value where the function is decreasing, but concave down.
Preview  

(d) Give an  x
 value where the function is increasing, but concave up.
Preview  

(e) Give an  x
 value where the function is decreasing, but concave up.
Preview  

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. 




Consider the function f(x) = 1 / (x^4 + 1)^2.

(a) Give the exact x value of a non-stationary inflection point (an inflection point which is not a critical point).

(b) Give an x value where the function is increasing, but concave down.

(c) Give an x value where the function is decreasing, but concave down.

(d) Give an x value where the function is increasing, but concave up.

(e) Give an x value where the function is decreasing, but concave up.

Explore the concavity, critical points, and inflection points of the function f(x) = 1 / (x^4 + 1)^2 through detailed calculus solutions, including the first and second derivatives. This problem involves concepts of critical points, concavity, and inflection points. Ideal for college-level calculus students.

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