The function f (x) = 1 – x³ – x5 is decreasing for

The correct option is all values of x.

To determine where a function is decreasing, we analyze the sign of its first derivative. A function $f\left(x\right)$ is strictly decreasing on an interval if its derivative $f^{\prime }\left(x\right)<0$ for all $x$ in that interval (except possibly at isolated points where the derivative can be zero).

Let the given function be:
$f\left(x\right)=1-x^3-x^5$

Now, we differentiate $f\left(x\right)$ with respect to $x$ using the power rule of differentiation:
$f^{\prime }\left(x\right)=\frac{d}{dx}\left(1-x^3-x^5\right)$
$f^{\prime }\left(x\right)=0-3x^2-5x^4$
$f^{\prime }\left(x\right)=-\left(3x^2+5x^4\right)$

Let's analyze the term inside the parentheses, $3x^2+5x^4$:
Since the exponents of $x$ are even (2 and 4), $x^2\ge 0$ and $x^4\ge 0$ for all real numbers $x$.
Consequently, $3x^2+5x^4\ge 0$ for all values of $x$.
This expression only equals zero when $x=0$. For all non-zero values of $x$, we have $3x^2+5x^4>0$.

Applying the negative sign outside the parentheses, we get:
$f^{\prime }\left(x\right)=-\left(3x^2+5x^4\right)\le 0$ for all real values of $x$.

Specifically, $f^{\prime }\left(x\right)<0$ for all $x\ne 0$, and $f^{\prime }\left(0\right)=0$. Since the derivative is strictly negative everywhere except at a single isolated point ($x=0$), the function $f\left(x\right)$ is strictly decreasing on the entire set of real numbers.

Therefore, the function is decreasing for all values of x.