Question

f(x)= x sqrt(x + 4)

which intervals is it increasing and decreasing on?
which interval is it concave up?

Answer 1

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

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