When is $f=\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^5-3x+\ln5$ decreasing $\forall\; x$? When is $$f=\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^5-3x+\ln5$$ decreasing $\forall\; x$?

Diffrentiating:
$$f'=5\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^4-3$$
If $f$ is decreasing, $f'<0$:
$$5\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^4-3<0\implies x^4<\frac3{5\left(\frac{\sqrt{p+4}}{1-p}-1\right)}$$
Since if $\left(\frac{\sqrt{p+4}}{1-p}-1\right)\to0^+\implies x^4<+\infty$, which holds forallx, So:$$p=\frac{3\pm\sqrt{21}}2$$which doesn't ply with textbook-answer, since it mentions an domain for p.

In order the function be always decreasing, the requirement is that $$f'=5\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^4-3$$ be negative for any value of $x$. Since $x^4 \geq 0$, then $$\frac{\sqrt{p+4}}{1-p}-1 \lt 0$$ which implies either $p \gt 1$ or $-4 \lt p \lt \frac{1}{2} \left(3-\sqrt{21}\right)$
