## 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. mattt9

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

2. mathcalculus

yup

3. mathcalculus

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

4. mathcalculus

@mattt9

5. mattt9

and what was your derivative again?

6. mathcalculus

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

7. mattt9

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

8. mathcalculus

9. mathcalculus

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

10. mattt9

|dw:1363315804166:dw|

11. mattt9

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

12. mattt9

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

13. mathcalculus

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

14. mattt9

as a random example

15. mattt9

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

16. mathcalculus

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

17. mathcalculus

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

18. mathcalculus

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

19. mattt9

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

20. mathcalculus

awesome so let me do that.

21. mattt9

okay

22. mathcalculus

how did you get 4x^2-28?

23. mathcalculus

what happened with the -6x?

24. mattt9

what -6x

25. mathcalculus

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

26. mattt9

|dw:1363317471013:dw|

27. mattt9

that's the numerator

28. mathcalculus

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

29. mathcalculus

^**

30. mattt9

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

31. mattt9

that is the full derivative.

32. mattt9

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

33. mattt9

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

34. mattt9

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

35. mattt9

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

36. mattt9

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.

37. mathcalculus

matt you make sense. are you there @mattt9

38. mathcalculus

so we don't test the critical points correct?

39. mattt9

those are the critical points.

40. mathcalculus

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

41. mattt9

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

42. mathcalculus

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...

43. mathcalculus

@mattt9 i would appreciate the help very much.

44. mattt9

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

45. mathcalculus

2.626 into the fn.?

46. mathcalculus

47. mattt9

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

48. mattt9

that should be your minimum for both.

49. mathcalculus

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

50. mathcalculus

thank you!!

51. mattt9

you areee welcome !