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

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- anonymous

do we use the quadratic formula for this one?

- anonymous

did you find the derivative?

- anonymous

derivative is 1*x^2+7-2x*(-4)*x/ (x^2+7)^2

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## More answers

- anonymous

from there.. x^2-6x+7/x^4+14x^2+49

- anonymous

then i = to 0

- anonymous

and i'm left with x^2-6x+7

- anonymous

however, i can't factor this... so do i use quadratic formula?

- anonymous

yeah I did not check your math, however if it is correct then use the quadratic formula.

- anonymous

can you try the problem also?

- anonymous

i want to know if its correct.

- zepdrix

"derivative is 1*x^2+7-2x*(-4)*x/ (x^2+7)^2"
That first term shouldn't be a 1.
The derivative of -4x is not 1.
I think that should fix it up for you.

- anonymous

yes i must of typed it wrong. but that's exactly what i got here.

- zepdrix

\[\large f'(x)=\frac{\color{royalblue}{(-4)}(x^2+7)-(-4x)\color{royalblue}{(2x)}}{(x^2+7)^2}\]

- anonymous

then after..

- zepdrix

Set it equal to zero, then multiply both sides by the denominator.
Don't do long division or anything silly like that.

- anonymous

yeah i ended up with x^2-6x+7/x^4+14x^2+49

- anonymous

??

- zepdrix

Sorry website wasn't working...

- zepdrix

\[\large 0=\frac{-4x^2-28+8x^2}{(x^2+7)^2}\]Multiplying both sides by the denominator gives us,\[\large 0=-4x^2-28+8x^2\]
I don't understand how you got the x term in the middle.

- anonymous

how do i get the derivative. can you show me?

- zepdrix

\[\large f(x)=\frac{-4x}{x^2+7}\]
Remember the quotient rule, it will tell us to setup the derivative like this,\[\large f'(x)=\frac{\color{royalblue}{(-4x)'}(x^2+7)-(-4x)\color{royalblue}{(x^2+7)'}}{(x^2+7)^2}\]The blue terms are the ones we need to differentiate.

- anonymous

yeah

- anonymous

according to the quotient rule.

- zepdrix

Which gives us this,\[\large f'(x)=\frac{\color{royalblue}{(-4)}(x^2+7)-(-4x)\color{royalblue}{(2x)}}{(x^2+7)^2}\]
Which simplifies to this,\[\large f'(x)=\frac{-4x^2-28+8x^2}{(x^2+7)^2}\]Right?

- anonymous

oh ok thank u!

- anonymous

then it's 4x^2-28 right?

- zepdrix

Yes. From there you can find a critical point.

- anonymous

radical + or -7

- anonymous

?

- anonymous

@zepdrix

- anonymous

i appreciate your help a lot. i just need time to figure this out... :(

- zepdrix

Yah that sounds right.
So we have a couple steps now.
We plug our `critical points` into the original function, and write down the \(\large f\) values they produce.
Then, plug the `end points` into the original function, and write down the \(\large f\) values they produce.
Then simply compare the \(\large f\) values. The largest will be your maximum. The smallest, your minimum.

- zepdrix

\[\large x \in\left[1,4\right]\]These are our end points, 1 and 4. The end points of our interval.

- anonymous

is the first critical point: -2 radical 7/7 ?

- anonymous

thats the positive criticla point that i used.

- anonymous

@zepdrix

- zepdrix

yah that sounds right.

- anonymous

and the f (- radical 7) is 4 radical 7?

- anonymous

hope it's right :(

- anonymous

are you checking to see if i am doing the problem correct?

- anonymous

very worried, please help me.

- zepdrix

no the negative root should produce 2sqrt7/7 i think.

- anonymous

?

- anonymous

it's not \[\frac{ -2\sqrt{7} }{ 7 }\] ??

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