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 ??

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

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

- anonymous

yup

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

@mattt9

- anonymous

and what was your derivative again?

- anonymous

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

- anonymous

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

- anonymous

yeah explain that please

- anonymous

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

- anonymous

|dw:1363315804166:dw|

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

as a random example

- anonymous

you wouldn't use the 7 when determining the derivative of the bottom since it is not attached to an x value.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

awesome so let me do that.

- anonymous

okay

- anonymous

how did you get 4x^2-28?

- anonymous

what happened with the -6x?

- anonymous

what -6x

- anonymous

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

- anonymous

|dw:1363317471013:dw|

- anonymous

that's the numerator

- anonymous

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

- anonymous

^**

- anonymous

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

- anonymous

that is the full derivative.

- anonymous

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

- anonymous

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

- anonymous

then solve for x to obtain a value for which the slope of the function is zero (a possible max or min)

- anonymous

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

- anonymous

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.

- anonymous

matt you make sense. are you there @mattt9

- anonymous

so we don't test the critical points correct?

- anonymous

those are the critical points.

- anonymous

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

- anonymous

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

- anonymous

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...

- anonymous

@mattt9 i would appreciate the help very much.

- anonymous

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

- anonymous

2.626 into the fn.?

- anonymous

the radical 7 or negative radical 7 is 2.64575

- anonymous

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

- anonymous

that should be your minimum for both.

- anonymous

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

- anonymous

thank you!!

- anonymous

you areee welcome !

Looking for something else?

Not the answer you are looking for? Search for more explanations.