Question

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 ??

Answer 1

do we use the quadratic formula for this one?

Answer 2

did you find the derivative?

Answer 3

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

Answer 4

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

Answer 5

then i = to 0

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

can you try the problem also?

Answer 10

i want to know if its correct.

Answer 11

"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.

Answer 12

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

Answer 13

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

Answer 14

then after..

Answer 15

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

Answer 16

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

Answer 17

??

Answer 18

Sorry website wasn't working...

Answer 19

\[\large 0=\frac{-4x^2-28+8x^2}{(x^2+7)^2}\]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.

Answer 20

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

Answer 21

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

Answer 22

yeah

Answer 23

according to the quotient rule.

Answer 24

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?

Answer 25

oh ok thank u!

Answer 26

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

Answer 27

Yes. From there you can find a critical point.

Answer 28

radical + or -7

Answer 29

?

Answer 30

@zepdrix

Answer 31

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

Answer 32

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.

Answer 33

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

Answer 34

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

Answer 35

thats the positive criticla point that i used.

Answer 36

@zepdrix

Answer 37

yah that sounds right.

Answer 38

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

Answer 39

hope it's right :(

Answer 40

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

Answer 41

very worried, please help me.

Answer 42

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

Answer 43

?

Answer 44

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