How do you find the intervals of increasing and decreasing using the first derivative given #y=xsqrt(16-x^2)#?

Question

1158 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-30T14:15:54

The interval of increasing is #(-sqrt8,sqrt8)#
The intervals of decreasing are #(-4,-sqrt8) uu (sqrt8,4)#

The function is #y=xsqrt(16-x^2)#

We need the domain of #y#

#16-x^2>=0#, #=>#, #x^2>=16#

Therefore the domain is #x in (-4,4)#

We need the derivative is

#(uv)'=u'v+uv'#

#u(x)=x#, #=>#, #u'(x)=1#

#v(x)=sqrt(16-x^2)#, #=>#, #v'(x)=(-2x)/(2sqrt(16-x^2))=-x/(sqrt(16-x^2))#

So,

#dy/dx=sqrt(16-x^2)-x^2/(sqrt(16-x^2))=(16-x^2-x^2)/(sqrt(16-x^2))#

#=(16-2x^2)/(sqrt(16-x^2))#

The critical points are when #dy/dx=0#

#(16-2x^2)/(sqrt(16-x^2))=0#

That is

#16-x^2=0#, #=>#, #x^2=8#, #=>#, #x=+-sqrt8#

We can build the variation chart

#color(white)(aaaa)##Interval##color(white)(aaaa)##(-4,-sqrt8)##color(white)(aaaa)##(-sqrt8, sqrt8)##color(white)(aaaa)##(sqrt8,4)#

#color(white)(aaaa)##dy/dx##color(white)(aaaaaaaaaaaaa)##-##color(white)(aaaaaaaaaaaa)##+##color(white)(aaaaaaaaa)##-#

#color(white)(aaaa)##y##color(white)(aaaaaaaaaaaaaaa)##↘##color(white)(aaaaaaaaaaaa)##↗##color(white)(aaaaaaaaa)##↘#

graph{xsqrt(16-x^2) [-16.02, 16.01, -8.01, 8.01]}