mathcalculus
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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mathcalculus
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do we use the quadratic formula for this one?
mattt9
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did you find the derivative?
mathcalculus
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derivative is 1*x^2+7-2x*(-4)*x/ (x^2+7)^2
mathcalculus
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from there.. x^2-6x+7/x^4+14x^2+49
mathcalculus
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then i = to 0
mathcalculus
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and i'm left with x^2-6x+7
mathcalculus
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however, i can't factor this... so do i use quadratic formula?
mattt9
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yeah I did not check your math, however if it is correct then use the quadratic formula.
mathcalculus
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can you try the problem also?
mathcalculus
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i want to know if its correct.
zepdrix
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"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.
mathcalculus
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yes i must of typed it wrong. but that's exactly what i got here.
zepdrix
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\[\large f'(x)=\frac{\color{royalblue}{(-4)}(x^2+7)-(-4x)\color{royalblue}{(2x)}}{(x^2+7)^2}\]
mathcalculus
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then after..
zepdrix
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Set it equal to zero, then multiply both sides by the denominator.
Don't do long division or anything silly like that.
mathcalculus
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yeah i ended up with x^2-6x+7/x^4+14x^2+49
mathcalculus
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??
zepdrix
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Sorry website wasn't working...
zepdrix
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\[\large 0=\frac{-4x^2-28+8x^2}{(x^2+7)^2}\]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.
mathcalculus
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how do i get the derivative. can you show me?
zepdrix
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\[\large f(x)=\frac{-4x}{x^2+7}\]
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.
mathcalculus
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yeah
mathcalculus
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according to the quotient rule.
zepdrix
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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?
mathcalculus
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oh ok thank u!
mathcalculus
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then it's 4x^2-28 right?
zepdrix
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Yes. From there you can find a critical point.
mathcalculus
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radical + or -7
mathcalculus
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?
mathcalculus
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@zepdrix
mathcalculus
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i appreciate your help a lot. i just need time to figure this out... :(
zepdrix
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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
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\[\large x \in\left[1,4\right]\]These are our end points, 1 and 4. The end points of our interval.
mathcalculus
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is the first critical point: -2 radical 7/7 ?
mathcalculus
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thats the positive criticla point that i used.
mathcalculus
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@zepdrix
zepdrix
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yah that sounds right.
mathcalculus
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and the f (- radical 7) is 4 radical 7?
mathcalculus
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hope it's right :(
mathcalculus
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are you checking to see if i am doing the problem correct?
mathcalculus
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very worried, please help me.
zepdrix
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no the negative root should produce 2sqrt7/7 i think.
mathcalculus
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?
mathcalculus
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it's not \[\frac{ -2\sqrt{7} }{ 7 }\] ??