The function f(x) = 1 - x^3 - x^5 is a decrea | vedclass

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(D) Given the function $f(x) = 1 - x^3 - x^5$. To determine where the function is decreasing,we find its derivative $f'(x)$. $f'(x) = \frac{d}{dx}(1 -

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$\forall x \in \mathbb{R}$

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(D) Given the function $f(x) = 1 - x^3 - x^5$. To determine where the function is decreasing,we find its derivative $f'(x)$. $f'(x) = \frac{d}{dx}(1 - x^3 - x^5) = -3x^2 - 5x^4$. We can factor out $-x^2$: $f'(x) = -x^2(3 + 5x^2)$. Since $x^2 \geq 0$ and $(3 + 5x^2) > 0$ for all $x \in \mathbb{R}$,it follows that $f'(x) = -x^2(3 + 5x^2) \leq 0$ for all $x \in \mathbb{R}$. Since $f'(x) \leq 0$ for all $x \in \mathbb{R}$,the function $f(x)$ is a decreasing function for all $x \in \mathbb{R}$.

The function $f(x) = 1 - x^3 - x^5$ is a decreasing function for:

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