## mathcalculus 2 years ago 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 ??

1. mathcalculus

do we use the quadratic formula for this one?

2. mattt9

did you find the derivative?

3. mathcalculus

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

4. mathcalculus

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

5. mathcalculus

then i = to 0

6. mathcalculus

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

7. mathcalculus

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

8. mattt9

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

9. mathcalculus

can you try the problem also?

10. mathcalculus

i want to know if its correct.

11. 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.

12. mathcalculus

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

13. zepdrix

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

14. mathcalculus

then after..

15. zepdrix

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

16. mathcalculus

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

17. mathcalculus

??

18. zepdrix

Sorry website wasn't working...

19. 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.

20. mathcalculus

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

21. 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.

22. mathcalculus

yeah

23. mathcalculus

according to the quotient rule.

24. 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?

25. mathcalculus

oh ok thank u!

26. mathcalculus

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

27. zepdrix

Yes. From there you can find a critical point.

28. mathcalculus

29. mathcalculus

?

30. mathcalculus

@zepdrix

31. mathcalculus

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

32. 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.

33. zepdrix

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

34. mathcalculus

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

35. mathcalculus

thats the positive criticla point that i used.

36. mathcalculus

@zepdrix

37. zepdrix

yah that sounds right.

38. mathcalculus

39. mathcalculus

hope it's right :(

40. mathcalculus

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

41. mathcalculus

42. zepdrix

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

43. mathcalculus

?

44. mathcalculus

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