IF `f(x)=1-x^3-x^5` is decreasing for

Options

A. `1lexle5`
B. `xle1`
C. `xge1`
D. All values of x

Answer

To determine the intervals where the function \( f(x) = 1 - x^3 - x^5 \) is decreasing, we need to find its derivative \( f'(x) \) and analyze where this derivative is less than zero.

### Step 1: Find the derivative of \( f(x) \)

The function is given by:
\[ f(x) = 1 - x^3 - x^5 \]

Now, we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}(1) - \frac{d}{dx}(x^3) - \frac{d}{dx}(x^5) \]
\[ f'(x) = 0 - 3x^2 - 5x^4 \]
\[ f'(x) = -3x^2 - 5x^4 \]

### Step 2: Set the derivative less than zero

To find where \( f(x) \) is decreasing, we set the derivative \( f'(x) \) less than zero:
\[ -3x^2 - 5x^4 < 0 \]

### Step 3: Factor the inequality

We can factor out the negative sign:
\[ -(3x^2 + 5x^4) < 0 \]
This implies:
\[ 3x^2 + 5x^4 > 0 \]

### Step 4: Analyze the factors

Notice that both \( 3x^2 \) and \( 5x^4 \) are always non-negative for all real values of \( x \) (since squares of real numbers are non-negative). The expression \( 3x^2 + 5x^4 \) will be zero only when \( x = 0 \).

### Step 5: Conclusion

Since \( 3x^2 + 5x^4 > 0 \) for all \( x \neq 0 \), we conclude that:
- \( f'(x) < 0 \) for all \( x \neq 0 \)
- Therefore, \( f(x) \) is decreasing for all \( x \in \mathbb{R} \) except at \( x = 0 \).

### Final Answer

The function \( f(x) = 1 - x^3 - x^5 \) is decreasing for all \( x \in \mathbb{R} \).

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