Find the inlerwatk on which $$ f(x) $$ s ncroasing and tho intervels on which $$ f(x) $$ is decreasing. Thon skelch the graph. Add horizontal tangert lines: $$ f(x)=2 x^{4}-100 x^{2} $$ Compcto the donvativo of I(x). $$ I^{\prime}(x)=8 x^{3}-200 x $$ Select the coried chcica holow and, if necessary, fill in the answer box to complale your choice. A. The funclion is increasing on $$ \square $$ (Typo your answor using intorval notation. Uso a comma to seporals answers as needod.) B. The function is never increasing.

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To determine the intervals on which the function $$ f(x) = 2x^4 - 100x^2 $$ is increasing or decreasing, we first need to analyze its derivative $$ I'(x) = 8x^3 - 200x $$.

### Step 1: Find Critical Points
To find the critical points, we set the derivative equal to zero:

$$
8x^3 - 200x = 0
$$

Factoring out the common term:

$$
8x(x^2 - 25) = 0
$$

This gives us:

$$
8x = 0 \quad \text{or} \quad x^2 - 25 = 0
$$

From $$ 8x = 0 $$, we get $$ x = 0 $$.

From $$ x^2 - 25 = 0 $$, we get:

$$
x^2 = 25 \implies x = 5 \quad \text{or} \quad x = -5
$$

Thus, the critical points are $$ x = -5, 0, 5 $$.

### Step 2: Test Intervals
Next, we will test the sign of $$ I'(x) $$ in the intervals determined by these critical points: $$ (-\infty, -5) $$, $$ (-5, 0) $$, $$ (0, 5) $$, and $$ (5, \infty) $$.

1. **Interval $$ (-\infty, -5) $$**: Choose $$ x = -6 $$
   $$
   I'(-6) = 8(-6)^3 - 200(-6) = 8(-216) + 1200 = -1728 + 1200 = -528 \quad (\text{decreasing})
   $$

2. **Interval $$ (-5, 0) $$**: Choose $$ x = -1 $$
   $$
   I'(-1) = 8(-1)^3 - 200(-1) = 8(-1) + 200 = -8 + 200 = 192 \quad (\text{increasing})
   $$

3. **Interval $$ (0, 5) $$**: Choose $$ x = 1 $$
   $$
   I'(1) = 8(1)^3 - 200(1) = 8 - 200 = -192 \quad (\text{decreasing})
   $$

4. **Interval $$ (5, \infty) $$**: Choose $$ x = 6 $$
   $$
   I'(6) = 8(6)^3 - 200(6) = 8(216) - 1200 = 1728 - 1200 = 528 \quad (\text{increasing})
   $$

### Step 3: Summarize Results
- The function is **increasing** on the intervals $$ (-5, 0) $$ and $$ (5, \infty) $$.
- The function is **decreasing** on the intervals $$ (-\infty, -5) $$ and $$ (0, 5) $$.

### Step 4: Fill in the Answers
A. The function is increasing on $$ (-5, 0) \cup (5, \infty) $$.

B. The function is never increasing. (This statement is incorrect based on our findings.)

### Step 5: Sketch the Graph
Now, let's sketch the graph of $$ f(x) $$ and add horizontal tangent lines at the critical points.

1. **Critical Points**: $$ f(-5) $$, $$ f(0) $$, and $$ f(5) $$.
2. **Horizontal Tangent Lines**: At $$ x = -5 $$, $$ x = 0 $$, and $$ x = 5 $$.

Let's calculate the values of $$ f(x) $$ at these critical points:

$$
f(-5) = 2(-5)^4 - 100(-5)^2 = 2(625) - 100(25) = 1250 - 2500 = -1250
$$
$$
f(0) = 2(0)^4 - 100(0)^2 = 0
$$
$$
f(5) = 2(5)^4 - 100(5)^2 = 2(625) - 100(25) = 1250 - 2500 = -1250
$$

Now, we can plot the graph and add horizontal tangent lines at these points.

### Final Answer
A. The function is increasing on $$ (-5, 0) \cup (5, \infty) $$.
A. The function is increasing on $$ (-5, 0) \cup (5, \infty) $$.

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