if f is negative then f is negative concave down cannot be determined decreasing

Answer:

D. decreasing

Explanation:

Step1: Recall derivative - function relationship

The sign of the first - derivative $f^{\prime}(x)$ of a function $y = f(x)$ determines the monotonicity of the function. If $f^{\prime}(x)>0$ for all $x$ in an interval $I$, then $y = f(x)$ is increasing on the interval $I$. If $f^{\prime}(x)<0$ for all $x$ in an interval $I$, then $y = f(x)$ is decreasing on the interval $I$.

Step2: Analyze the given condition

We are given that $f^{\prime}$ is negative. By the above - mentioned rule, when $f^{\prime}(x)<0$ for all $x$ in the domain of consideration, the function $f(x)$ is decreasing. The sign of $f^{\prime}$ does not directly tell us about the sign of $f$ (a function can be positive while its derivative is negative, for example, $f(x)= - x + 2$ has $f^{\prime}(x)=-1<0$ and $f(0) = 2>0$). The sign of the second - derivative $f^{\prime\prime}$ determines the concavity of the function. If $f^{\prime\prime}(x)<0$, the function is concave down, and we are only given information about $f^{\prime}$, not $f^{\prime\prime}$. So, we can determine that if $f^{\prime}$ is negative, $f$ is decreasing.