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 √(x+4) \\ \\ F'(x)=(x √(x+4))'=√(x+4)+ (x)/(2 √(x+4) ) \\ \\F'(x)=0 \\ \\ √(x+4)+ (x)/(2 √(x+4) ) =0 \\ \\ (2(x+4))/(2 √(x+4) )+(x)/(2 √(x+4) ) =0 \\ \\(3x+8)/(2 √(x+4) ) =0 \\ \\3x+8=0 \\x=- (8)/(3) \\ x\in(-\infty,- (8)/(3) )\Rightarrow y\downarrow \\ x\in(- (8)/(3),+\infty )\Rightarrow y\uparrow`


`F''(x)=((3x+8)/(2 √(x+4) ))'= (3*2√(x+4)- (3x+8)/( √(x+4) ) )/(4(x+4)) = (6(x+4)-(3x+8))/(4(x+4)√(x+4)) = (3x+16)/(4(x+4)√(x+4)) \\ \\F''(x)=0 \\ \\ (3x+16)/(4(x+4)√(x+4)) =0 \\ \\3x+16=0 \\ \\x=- (16)/(3) \\ \\x\in(-\infty,- (16)/(3))\cup(-4,+\infty)\Rightarrow F''(x) > 0\Rightarrow \text{ F(x) concave up}`