The function f(x) = 1 - x^3 - x^5 is decreasi | 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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All values of $x$

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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 $-x^2$ from the expression: $f'(x) = -x^2(3 + 5x^2)$. Since $x^2 \ge 0$ for all real $x$ and $(3 + 5x^2) > 0$ for all real $x$,the product $-x^2(3 + 5x^2)$ is always less than or equal to $0$. Specifically,$f'(x) Since the derivative is non-positive for all $x$ and does not vanish on any interval,the function $f(x)$ is strictly decreasing for all real values of $x$.

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

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