Find the value of p for which the function f(x)=$\left(\frac{\sqrt{p+4}}{1p}1\right)x^{5}3x+ln5$ decreases $\forall$x $\in\mathbb{R}$
we have to solve $$\frac{\sqrt{p+4}}{1p}\le 1$$ if $1p<0$ so $1<p$ then our inequality is fulfilled. no we assume $1p>0$ then we have to solve $$0\le p^23p3\geq 0$$ this gives us $$\frac{3}{2}+\frac{1}{2}\sqrt{21}\le p<+\infty$$ or $$\infty<p\le \frac{3}{2}\frac{1}{2}\sqrt{21}$$ as you stated
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Kislay Tripathi
Updated on August 19, 2022Comments

Kislay Tripathi about 1 year
Question Find the value of p for which the function f(x)=$\left(\frac{\sqrt{p+4}}{1p}1\right)x^{5}3x+ln5$ decreases $\forall$x $\in\mathbb{R}$
MY Approach$\Longrightarrow$f'$\left(x\right)$= $\left(\frac{\sqrt{p+4}}{1p}1\right)5x^{4}3$$\leq$0$\Longrightarrow$$\left(\frac{\sqrt{p+4}}{1p}1\right)\leq$$\frac{3}{5x^{4}}$$\Longrightarrow$$\left(\frac{\sqrt{p+4}}{1p}1\right)<0$$\Longrightarrow$x $\in$$\left(\infty,\frac{3\sqrt{21}}{2}\right)$$\bigcup$ $\left(\frac{3+\sqrt{21}}{2},\infty\right)$
But Book Says Answer is $\left[4,\frac{3\sqrt{21}}{2}\right]$$\bigcup$$\left(1,\infty\right)$

Kislay Tripathi about 6 years$$\frac{\sqrt{p+4}}{1p}\le 1$$ if i square this equation i do not get the condition x belongs in (1,0) but $$\frac{\sqrt{p+4}}{1}\le1p $$ this eqaution gives the condition x belongs in (1,0).How to know the correct approach