Question

find the critical points of y=(x^2)/(4x+4)

Answer 1

Do you know what you have to do?

Answer 2

take the derivative, and set it equal to 0

Answer 3

Yup.

Answer 4

Can you find the derivative of that?

Answer 5

u'v - uv' / v^2

Remember the denominator is u and the numerator is v.

Answer 6

yes, i believe it is (4x^2+8)/(4x+4)^2

Answer 7

looks almost right...

\[\large y'=\frac{ (4x+4)(2x)-x^2(4) }{ (4x+4)^2 }=\frac{ 4x^2+8x }{ (4x+4)^2 }\]

Answer 8

\[\large = \frac{ 4x(x+2) }{ 16(x+1)^2 }=\frac{ x(x+2) }{ 4(x+1)^2 }\]

Answer 9

so the points would be -2,-1, and 0??

Answer 10

right, the x values to test

Answer 11

ok thanks :)

Answer 12

test to see which ones if any are crit points

Answer 13

oh an um do we seperatley equal them to 0? for example x(x+2)=0 or 4(x+1)^2
or do we do it as a whole?

Answer 14

A Critical Point at x=c says,
f(c) exists
f '(c)=0 or f '(c) doesnt exist

Answer 15

The original function has x=-1 not existing. All other values are fine for x. So only the two points
(-2 , -1) and (0,0) are critical points