Mathematics: F(x)= x sqrt(x + 4) which intervals is it increasing and decreasing on? which interval is it concave up?

F(x)= x sqrt(x + 4) which intervals is it increasing

Ответ:

cluchmasters3831

03.01.2022

$F(x)=x \sqrt{x+4}
\\
\\ F'(x)=(x \sqrt{x+4})'=\sqrt{x+4}+ \frac{x}{2 \sqrt{x+4} } 
\\
\\F'(x)=0
\\
\\ \sqrt{x+4}+ \frac{x}{2 \sqrt{x+4} } =0
\\
\\ \frac{2(x+4)}{2 \sqrt{x+4} }+\frac{x}{2 \sqrt{x+4} } =0
\\
\\\frac{3x+8}{2 \sqrt{x+4} } =0
\\
\\3x+8=0
\\x=- \frac{8}{3} 
\\ x\in(-\infty,- \frac{8}{3} )\Rightarrow y\downarrow
\\ x\in(- \frac{8}{3},+\infty )\Rightarrow y\uparrow$

$F''(x)=(\frac{3x+8}{2 \sqrt{x+4} })'= \frac{3\times2\sqrt{x+4}- \frac{3x+8}{ \sqrt{x+4} } }{4(x+4)} = \frac{6(x+4)-(3x+8)}{4(x+4)\sqrt{x+4}} =
 \frac{3x+16}{4(x+4)\sqrt{x+4}} 
\\
\\F''(x)=0
\\
\\ \frac{3x+16}{4(x+4)\sqrt{x+4}} =0
\\
\\3x+16=0
\\
\\x=- \frac{16}{3} 
\\
\\x\in(-\infty,- \frac{16}{3})\cup(-4,+\infty)\Rightarrow F''(x) 0\Rightarrow \text{ F(x) concave up}$