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is the x^2+7 all under the -4x

yup

yup\[ \frac{ (-4)*x}{ x^{2}+7}\]

and what was your derivative again?

\[\frac{ 1*(x ^{2}+ 7)-2x(-4)*x}{ (x^{2}+7) ^2 }\]

are you sure that is correct wouldn't the derivative of the top be -4 not 1

yeah explain that please

maybe I did get it wrong. but i assumed since -4 is a constant i didn't use it...

|dw:1363315804166:dw|

you still use the constant if it's attached to the x like that.

(d/dx) of 2x is 2 right?

oh so when you add you don't use the constant right?

as a random example

oh so you multiplied them which is -4x... then just fnd the derivative right>

for example if it is -4+x then the derivative is 1.

but if (-4)*x you keep them as -4 and find the derivative which is -4

right. then your derivative simplifies to :
\[\frac{ (4x^2-28) }{ (x^2+7)(x^2+7) }\]

awesome so let me do that.

okay

how did you get 4x^2-28?

what happened with the -6x?

what -6x

i thought the derivative would be -4x^2-28-6x on the numerator

|dw:1363317471013:dw|

that's the numerator

yeah but you wrote.. (4x2−28)/(x2+7)(x2+7)

^**

yeah i know. but you had problems with the numerator. so i just showed you how i got the numerator

that is the full derivative.

anyways. you end up equation that to zero.
\[\frac{ (4x^2-28) }{ (x^2+7)(x^2+7) }=0\]

the (x^2+7) cancels out and you are left with 4x^2-28 = 0

at this point you will see that x = +-root(7) = +- 2.646

so we don't test the critical points correct?

those are the critical points.

a) because the intervals are [1,4] and b) intervals are from [1,5]

ummmmm what do you get when you plug in 2.626 into the fn.?

2.626 into the fn.?

the radical 7 or negative radical 7 is 2.64575

(-4)(2.626)/(2.626^2+7) = -0.7559

that should be your minimum for both.

youre right, i didn't see the negative by mistake.

thank you!!

you areee welcome !