22. The function $$ f(x)=-x^{4}+72 x^{2} $$ has $$ \begin{array}{llll} ext { (A. one relative minimum and two relative maxima. } \ ext { B. one relative minimum and one relative maximum. } \ ext { C. no relative minimum and two relative maxima. } \ ext { D. two relative minima and no relative maximum. } \ ext { E. two relative minima and one relative maximum. } \ \begin{array}{llll} ext { A. }-12,0,12 & ext { B. }-12,12 & ext { C. } \sqrt{12} & ext { D. }-\sqrt{12}, \sqrt{12}\end{array} \quad ext { E. }-\sqrt{12}, 0, \sqrt{12}\end{array} $$ $$ \begin{array}{lll} ext { 23. Find the } x ext {-coordinate(s) of the point(s) of inflection of the curve in problem } 22\end{array} $$

Question

22. The function $$ f(x)=-x^{4}+72 x^{2} $$ has $$ \begin{array}{llll}	ext { (A. one relative minimum and two relative maxima. } \ 	ext { B. one relative minimum and one relative maximum. } \ 	ext { C. no relative minimum and two relative maxima. } \ 	ext { D. two relative minima and no relative maximum. } \ 	ext { E. two relative minima and one relative maximum. } \ \begin{array}{llll}	ext { A. }-12,0,12 & 	ext { B. }-12,12 & 	ext { C. } \sqrt{12} & 	ext { D. }-\sqrt{12}, \sqrt{12}\end{array} \quad 	ext { E. }-\sqrt{12}, 0, \sqrt{12}\end{array} $$ $$ \begin{array}{lll}	ext { 23. Find the } x 	ext {-coordinate(s) of the point(s) of inflection of the curve in problem } 22\end{array} $$

UpStudy AI · Accepted Answer

Let's analyze the function $$ f(x) = -x^4 + 72x^2 $$ to determine its relative extrema and points of inflection.

### Problem 22: Relative Extrema

**Finding Critical Points:**
First, find the first derivative of $$ f(x) $$:
$$
f'(x) = \frac{d}{dx}(-x^4 + 72x^2) = -4x^3 + 144x = -4x(x^2 - 36) = -4x(x + 6)(x - 6)
$$
Setting $$ f'(x) = 0 $$ gives the critical points:
$$
x = 0, \quad x = 6, \quad x = -6
$$

**Determining Nature of Critical Points:**
Next, find the second derivative:
$$
f''(x) = \frac{d}{dx}(-4x^3 + 144x) = -12x^2 + 144 = -12(x^2 - 12)
$$
Evaluate $$ f''(x) $$ at each critical point:
- At $$ x = -6 $$:
  $$
  f''(-6) = -12((-6)^2) + 144 = -432 + 144 = -288 < 0 \quad \text{(Local Maximum)}
  $$
- At $$ x = 0 $$:
  $$
  f''(0) = -12(0)^2 + 144 = 144 > 0 \quad \text{(Local Minimum)}
  $$
- At $$ x = 6 $$:
  $$
  f''(6) = -12(6)^2 + 144 = -432 + 144 = -288 < 0 \quad \text{(Local Maximum)}
  $$

**Conclusion for Problem 22:**
- **One relative minimum at $$ x = 0 $$**
- **Two relative maxima at $$ x = -6 $$ and $$ x = 6 $$**

**Answer:** **A. one relative minimum and two relative maxima.**

### Problem 23: Points of Inflection

**Finding Points of Inflection:**
Points of inflection occur where the second derivative changes sign, i.e., where $$ f''(x) = 0 $$:
$$
-12x^2 + 144 = 0 \implies x^2 = 12 \implies x = \sqrt{12} \ \text{or} \ x = -\sqrt{12}
$$
Simplifying $$ \sqrt{12} $$:
$$
\sqrt{12} = 2\sqrt{3} \approx 3.464
$$

**Conclusion for Problem 23:**
- **Points of inflection at $$ x = -\sqrt{12} $$ and $$ x = \sqrt{12} $$**

**Answer:** **D. $$ -\sqrt{12} $$, $$ \sqrt{12} $$**
**Problem 22:** - **One relative minimum at $$ x = 0 $$** - **Two relative maxima at $$ x = -6 $$ and $$ x = 6 $$** **Answer:** **A. one relative minimum and two relative maxima.** **Problem 23:** - **Points of inflection at $$ x = -\sqrt{12} $$ and $$ x = \sqrt{12} $$** **Answer:** **D. $$ -\sqrt{12} $$, $$ \sqrt{12} $$**

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