$P(-1)$ is the minimum but $P(1)$ is not the maximum of $P$

$P(-1)$ is not minimum but $P(1)$ is the maximum of $P$

neither $P(1)$ is the minimum nor $P(1)$ is the maximum of P

$P(-1)$ is the minimum and $P(1)$ is the maximum of $P$.

If $f(x)$ is decreasing, then $\frac{1}{f(x)}$ is increasing

If $f(x)$ is decreasing, then $\frac{1}{f(x)}$ is also decreasing

If both $f$ and $g$ are positive functions such that $f$ is decreasing and $g$ is increasing, then $\frac{f}{g}$ is a decreasing function

If both $f$ and $g$ are positive functions such that $f$ is increasing and $g$ is decreasing then $\frac{f}{g}$ is a decreasing furnction

$p(x)=-3$

$\mathrm{p}(\mathrm{x})=0$

$0< p(x)<-h(4)$

$0 \leq p(x) \leq-h(4)$

$(1,2)$

$(-1,2)$

$\left(\frac{1}{2}, \frac{5}{4}\right)$

$(0,1)$