Find the values of 'x' for which the rate of change of `x^4/4+x^3/3-x` is more than `x^4/4`

Answer

To find the values of 'x' for which the rate of change of the function \( y = \frac{x^4}{4} + \frac{x^3}{3} - x \) is more than \( g = \frac{x^4}{4} \), we will follow these steps:

### Step 1: Define the functions
Let:
\[ y = \frac{x^4}{4} + \frac{x^3}{3} - x \]
\[ g = \frac{x^4}{4} \]

### Step 2: Find the derivatives
We need to find the derivatives of both functions with respect to \( x \).

1. Derivative of \( y \):
\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{x^4}{4}\right) + \frac{d}{dx}\left(\frac{x^3}{3}\right) - \frac{d}{dx}(x) \]
\[ = x^3 + x^2 - 1 \]

2. Derivative of \( g \):
\[ \frac{dg}{dx} = \frac{d}{dx}\left(\frac{x^4}{4}\right) = x^3 \]

### Step 3: Set up the inequality
We need to find when the rate of change of \( y \) is greater than the rate of change of \( g \):
\[ \frac{dy}{dx} > \frac{dg}{dx} \]
This translates to:
\[ x^3 + x^2 - 1 > x^3 \]

### Step 4: Simplify the inequality
Subtract \( x^3 \) from both sides:
\[ x^2 - 1 > 0 \]

### Step 5: Factor the inequality
Factoring gives:
\[ (x - 1)(x + 1) > 0 \]

### Step 6: Analyze the sign of the product
To find the intervals where this product is positive, we can find the critical points:
- The critical points are \( x = -1 \) and \( x = 1 \).

### Step 7: Test intervals
We will test the intervals determined by these critical points:
1. For \( x < -1 \) (e.g., \( x = -2 \)):
\[ (-2 - 1)(-2 + 1) = (-3)(-1) = 3 > 0 \quad \text{(positive)} \]
2. For \( -1 < x < 1 \) (e.g., \( x = 0 \)):
\[ (0 - 1)(0 + 1) = (-1)(1) = -1 < 0 \quad \text{(negative)} \]
3. For \( x > 1 \) (e.g., \( x = 2 \)):
\[ (2 - 1)(2 + 1) = (1)(3) = 3 > 0 \quad \text{(positive)} \]

### Step 8: Conclusion
The product \( (x - 1)(x + 1) > 0 \) is satisfied in the intervals:
\[ (-\infty, -1) \cup (1, \infty) \]

Thus, the values of \( x \) for which the rate of change of \( y \) is more than \( g \) are:
\[ x \in (-\infty, -1) \cup (1, \infty) \]