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Set of all values of '$a$' for which $f(x)=\bigg(\dfrac{\sqrt{a+4}}{1-a}-1\bigg)x^5-3x+\log 5$ decreases for all real $x$

$$ a\neq1\;\&\; a\geq -4\\ f'(x)=5\bigg(\dfrac{\sqrt{a+4}}{1-a}-1\bigg)x^4-3<0\\ \bigg(\dfrac{\sqrt{a+4}}{1-a}-1\bigg)x^4<\frac{3}{5}\text{ for all} x\in\mathcal{R}\\ \dfrac{\sqrt{a+4}}{1-a}-1=0\implies\sqrt{a+4}=1-a\\ a+4=1+a^2-2a\implies a^2-3a-3=0\implies a=\frac{3\pm\sqrt{21}}{2} $$

My reference gives the solution $a\in\big[-4,\dfrac{3-\sqrt{21}}{2}\big]\cup(1,\infty)$, yet I am stuck here. Could anyone give a hint of whats the way forward ?

Martin Sleziak
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Sooraj S
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2 Answers2

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reference gives wrong solution (e.g. try $a=-3$ and check the chart).

You had the right idea about checking the derivative. $f$ is differentiable, thus $f$ is decreasing equivalent to $f'(x) < 0$, which implies inequality you had:

$$\frac{\sqrt{a+4}}{1-a}x^4 < \frac{3}{5}$$

now notice that if $\frac{\sqrt{a+4}}{1-a}$ is positive, then for all $ x > (\frac{3}{5} \frac{1-a}{\sqrt{a+4}})^{\frac{1}{4}}$, inequality will not hold, thus we should be looking only for those $a$, for which $\frac{\sqrt{a+4}}{1-a} \le 0.$

if $\frac{\sqrt{a+4}}{1-a} = 0$, then condition holds and that gives you $a=-4$.

if $\frac{\sqrt{a+4}}{1-a} < 0$, then $\frac{\sqrt{a+4}}{1-a}x^4 < 0 < \frac{3}{5}$ - condition also holds and that gives you $a > 1$

the final answer: $\{-4\} ∪ \{1, \infty\}$

govordovsky
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Let $y=k x^5 -3x + \log 5 \implies y'=5kx^4 -3$ if $y(x)$ is increasing for all real values of $x$, then $y'=5kx^4 \le 3 \implies \frac{6k}{3} x^4 \le 0\implies k \le 0 .$ $$\implies \frac{\sqrt{a+4}}{1-a} \le 1 \implies a\ge -4.~~~ (1)$$ Additionally, two cases arise here:

Case 1: when $a < 1$: squaring (1), we get $a^2-3a-3 \ge 0 \implies a \in[-4, \frac{3-\sqrt{21}}{2}].$

Case 2: $a \in (1, \infty)$, then $\sqrt{a+4} \ge 1-a$, is true.

Finally, $a \in [-4,\frac{3-\sqrt{21}}{2}] \cup (1,\infty)$

Z Ahmed
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