Will the function $$ f(x)=\frac{12 x^{3}}{x^{2}-5 \dot{x}} $$ be changing more rapidly at $$ x=1 $$. or $$ x=-4 $$ ? Why? Use the paperclip button below to attach files.

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To determine whether the function $$ f(x) = \frac{12x^{3}}{x^{2} - 5x} $$ is changing more rapidly at $$ x = 1 $$ or $$ x = -4 $$, we'll analyze its rate of change by computing its derivative and evaluating it at these points.

### Step 1: Simplify the Function

First, simplify the denominator:

$$
f(x) = \frac{12x^{3}}{x^{2} - 5x} = \frac{12x^{3}}{x(x - 5)}
$$

### Step 2: Compute the Derivative

We'll use the quotient rule for differentiation, which states:

$$
\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}
$$

Let $$ u = 12x^{3} $$ and $$ v = x^{2} - 5x $$. Then:

$$
u' = 36x^{2}
$$
$$
v' = 2x - 5
$$

Applying the quotient rule:

$$
f'(x) = \frac{36x^{2}(x^{2} - 5x) - 12x^{3}(2x - 5)}{(x^{2} - 5x)^2}
$$

Simplify the numerator:

$$
36x^{2}(x^{2} - 5x) = 36x^{4} - 180x^{3}
$$
$$
12x^{3}(2x - 5) = 24x^{4} - 60x^{3}
$$

Subtracting the second part from the first:

$$
36x^{4} - 180x^{3} - (24x^{4} - 60x^{3}) = 12x^{4} - 120x^{3}
$$

Factor out common terms:

$$
12x^{3}(x - 10)
$$

So, the derivative simplifies to:

$$
f'(x) = \frac{12x^{3}(x - 10)}{(x(x - 5))^2} = \frac{12x(x - 10)}{(x - 5)^2}
$$

### Step 3: Evaluate the Derivative at $$ x = 1 $$ and $$ x = -4 $$

**At $$ x = 1 $$:**

$$
f'(1) = \frac{12 \times 1 \times (1 - 10)}{(1 - 5)^2} = \frac{12 \times (-9)}{16} = \frac{-108}{16} = -6.75
$$

**At $$ x = -4 $$:**

$$
f'(-4) = \frac{12 \times (-4) \times (-4 - 10)}{(-4 - 5)^2} = \frac{12 \times (-4) \times (-14)}{81} = \frac{672}{81} \approx 8.296
$$

### Step 4: Compare the Rates of Change

- **At $$ x = 1 $$:** The derivative is **-6.75**.
- **At $$ x = -4 $$:** The derivative is approximately **8.296**.

Since the **magnitude** of the derivative at $$ x = -4 $$ (approximately 8.296) is greater than that at $$ x = 1 $$ (6.75), the function $$ f(x) $$ is **changing more rapidly at $$ x = -4 $$**.

### Conclusion

The function $$ f(x) = \frac{12x^{3}}{x^{2} - 5x} $$ is changing more rapidly at $$ x = -4 $$ compared to $$ x = 1 $$ because the absolute value of the derivative at $$ x = -4 $$ is greater than at $$ x = 1 $$.
The function $$ f(x) = \frac{12x^{3}}{x^{2} - 5x} $$ changes more rapidly at $$ x = -4 $$ than at $$ x = 1 $$ because the derivative at $$ x = -4 $$ is larger in magnitude.

Will the function \( f(x)=\frac{12 x^{3}}{x^{2}-5 \dot{x}} \) be changing more rapidly at \( x=1 \). or \( x=-4 \) ? Why? Use the paperclip button below to attach files.

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