help help help!!!! Determine the extrema of f(x)= (-4)* x/ x^2+7 below on the given interval (a) on [1,4] The minimum is ?? and the maximum is ?? (b) on [1,5] The minimum is ?? and the maximum is ??

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help help help!!!! Determine the extrema of f(x)= (-4)* x/ x^2+7 below on the given interval (a) on [1,4] The minimum is ?? and the maximum is ?? (b) on [1,5] The minimum is ?? and the maximum is ??

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

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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
you wouldn't use the 7 when determining the derivative of the bottom since it is not attached to an x value.
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
then solve for x to obtain a value for which the slope of the function is zero (a possible max or min)
at this point you will see that x = +-root(7) = +- 2.646
then you see that x = -2.626 is not in your given interval so you would not test this value. in the first interval of [1,4] you would test the x-values of x = 1 x = 4 and your found x value of x = 2.626 out of these three numbers one will be highest and one will be lowest, those should be your max and min for the function respectively. Let me know if any of that is incorrect.
matt you make sense. are you there @mattt9
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]
or at least 2.626 is for the function. 1 and 4 are for the interval. or maybe that terminology isn't used i can't remember but you would test all three of those points
i did test them all out. i wrote (a) on [1,4] The minimum is -.695652 and the maximum is -1/2 (b) on [1,5] The minimum is -.625 and the maximum is -1/2 but when i click submit it says one of them is wrong...
@mattt9 i would appreciate the help very much.
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 !

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