## anonymous one year ago NEED HELP! What are the intervals of increase and decrease of f(x) = x ( sqrt of x+3) * This is a calc problem

1. freckles

Have you considered finding the derivative?

2. UsukiDoll

we need the first derivative test... wait we need to find critical points first!

3. freckles

f'>0 tells us f is increasing f'<0 tells us f is decreasing Find the critical numbers and consider the domain.

4. UsukiDoll

what @freckles said is correct... use product rule

5. anonymous

I already did all that, and I got (sqrt x+3) + x/(2(sqrt x +3)). But I just dont know how to go about solving for the critical values

6. freckles

I don't think we need to find the x-intercepts to find where the function is increasing/decreasing.

7. anonymous

is it just x= -3?

8. UsukiDoll

noooooo....................... freckles we need the critical points first or we can't go further T_T

9. UsukiDoll

sorry mary that comment was at freckles

10. freckles

$f'(x)=\sqrt{x+3}+\frac{x}{2 \sqrt{x+3}} \\ \text{ solve } f'(x)=0 \text{ \to find the critical numbers }$

11. Zale101

To get the critical points, we first need to set the derivative to zero

12. anonymous

Yea I did that already guys!!!

13. Zale101

and solve for x. Those x's are the crtical points.

14. freckles

good then what did you for the x's?

15. anonymous

I just need help in solving for x

16. freckles

$\sqrt{x+3} \frac{2\sqrt{x+3}}{2\sqrt{x+3}}+\frac{x}{2 \sqrt{x+3}}$ find common denominator

17. anonymous

No, we find the first derivative and then find the critical points

18. freckles

$\frac{2 (x+3)+x}{2 \sqrt{x+3}}=0$

19. anonymous

hey freckles, how did you get +x in the numerator?

20. freckles

(sqrt x+3) + x/(2(sqrt x +3) isn't this what you got for f'?

21. anonymous

OH, K i see it

22. freckles

$f'(x)=\sqrt{x+3}+\frac{x}{ 2 \sqrt{x+3}} \\ f'(x) =\frac{ \sqrt{x+3} }{1} \cdot \frac{2 \sqrt{x+3}}{2 \sqrt{x+3}} +\frac{x}{2 \sqrt{x+3}} \\ f'(x)=\frac{2(\sqrt{x+3})^2+x}{2 \sqrt{x+3}}$

23. anonymous

So is it x=-2 for when f'(x) = 0

24. freckles

yes

25. anonymous

and then the denominator would give me when derivative is undefined?

26. anonymous

so the critical points are x= -2, and -3?

27. freckles

in by the way the domain of f is x>=-3 so we have this so far: |dw:1437451331355:dw|

28. freckles

x=-3 is an endpoint

29. freckles

since the function doesn't exist to the left of x=-3

30. freckles

so you only have two intervals to consider the ones in the picture above just choose a number from each and plug into f'(x)

31. UsukiDoll

what about the interval (-oo, -3) . Don't we have to pick a number in that interval too?

32. freckles

the function doesn't exist to the left of x=-3

33. freckles

the domain of x *sqrt(x+3) is x+3>=0 because of that square root function

34. UsukiDoll

yeah and we can't have x = -3. I can see that part, but how did you guys get x = -2 ?

35. freckles

well we can have x=-3 but we can't have anything less than x=-3 and he and she and I got x=-2 from finding the critical numbers

36. freckles

The critical numbers are the numbers that satisfy either f'(x)=0 or f'(x) does not exist as long as those x values are in the domain of the original function f(x).

37. anonymous

why cant we say the interval is from (-3, Infinity)

38. freckles

remember @mary_am you found x=-2 as a critical number so we have to consider the intervals (-3,-2) and (-2,infty) separately

39. anonymous

gotcha, I think im overthinking it

40. freckles

choose a number from each of the intervals (not an endpoint)

41. freckles

plug into f'(x)

42. anonymous

so its increasing not decreasing on those two diff intervals, right?

43. UsukiDoll

$\frac{2 (x+3)+x}{2 \sqrt{x+3}}=0$ $\frac{ 3x+6}{2 \sqrt{x+3}}=0$ $\frac{ 3(x+2)}{2 \sqrt{x+3}}=0$ x=-3, -2 our intervals are based on the critical points (-3,-2) and (-2,oo) then pick a number between those intervals and plug it into f'(x)

44. anonymous

ya plugged the numbers in and got increasing for both those intervals, but just wanted to make sure I was correct

45. UsukiDoll

if it's negative <- decreasing if it's positive <- increasing

46. anonymous

and also would there be any local max or min values for this problem?

47. anonymous

because if its just increasing....I would believe that x=-3 would just be the local min...? but no so sure actually

48. UsukiDoll

we'll you were asked to find just increasing and decreasing so finding local max and local max isn't required.

49. UsukiDoll

which is the first derivative test

50. anonymous

lol the second part of the questions asks for the local max and min

51. UsukiDoll

I don't even see that you've typed that.

52. Zale101

You can find local min and max after you find what intervals are decreasing or increasing.

53. anonymous

yea I just did that and I got increasing for both invtervals and now im like super confused

54. Zale101

|dw:1437451995489:dw| If the graph goes from increasing to decreasing, then that point can be a local max, if it goes from decreasing to increasing then it is local min.

55. freckles

$f'(x)=\frac{3x+6}{2 \sqrt{x+3}} \\ \text{ a number from } (-2,-3) \text{ is } -2.5 \\ f'(-2.5)=\frac{3(-2.5)+6}{2 \sqrt{-2.5+3}}=\frac{-6.5+6}{2 \sqrt{.5}}$ is this positive or negative @mary_am ?

56. anonymous

should be negative

57. freckles

oops I meant (-3,-2)

58. freckles

but yes f'<0 on (-3,-2)

59. freckles

f'<0 means f' is negative

60. freckles

if f' is negative on (-3,-2) then f is increasing/decreasing on (-3,-2)?

61. anonymous

Ifound my mistake, I plugged in 0...

62. freckles

yeah 0 cannot be chosen from (-3,-2)

63. anonymous

lol yea....

64. UsukiDoll

for (-3,-2) we can only choose a very small selection

65. UsukiDoll

x = 0 is more appropriate in (-2,oo) though

66. freckles

you could choose some of the following: -2.99 -2.88 -2.8 -2.7 -2.6 -2.4 -2.1 any many more...

67. anonymous

okay, so its increasing from -2, infitnity and decreasing from (-3,-2), right?

68. anonymous

Yea I know I just dont know what happened... brain isnt functioning to properly

69. UsukiDoll

mine isn't either . It's in force to be awake mode.

70. freckles

ok sounds great to me except I would say on not from increasing on (-2,infty) decreasing on (-3,-2)

71. freckles

|dw:1437452317669:dw| so the function may look sorta like this I just used the information from f' and the domain of the original function to help me draw a rough graph of f

72. freckles

|dw:1437452372080:dw|

73. freckles

reading the graph from left to right of course

74. anonymous

But I am right when I say that the fucntion inreases from -2 to infinity and decreases from -3 to -2, right? and that there is a local minumum at x=-2

75. freckles

yes but that local min can also be a ?

76. UsukiDoll

it's backwards I would write, our function is decreasing from (-3,-2) and increasing from (-2,oo)

77. anonymous

OH it can also be an inflection point....?

78. freckles

Well I was going for global max inflection point you might want to consider when f''=0

79. anonymous

I also have to find the intervals for cancavity and inflection points

80. freckles

global min*

81. anonymous

so I know that Im supposed to the second derivative

82. anonymous

I've never heard of global min...

83. freckles

|dw:1437452577062:dw|

84. freckles

it is just the lowest point on the graph

85. UsukiDoll

for concavity take the derivative again... set that second derivative to 0 and find the critical points

86. anonymous

So is the second derivative = (x+3^-1/2) (2) - (2x+3)(-1/2x+3^-3/2)

87. anonymous

or am I like waaaay off....? :/

88. freckles

$f'(x)=\frac{3x+6}{2 \sqrt{x+3}} \\ f'(x)=\frac{(3x+6)' \cdot 2 \sqrt{x+3}-(2 \sqrt{x+3})' \cdot (3x+6)}{(2 \sqrt{x+3})^2}$

89. freckles

can you find (3x+6)' and (2 sqrt(x+3))'

90. anonymous

yea, hollup

91. anonymous

its 3 and the other one is x+3 ^-1/2

92. freckles

that is right $(3x+6)'=(3x)'+(6)'=3+0=3 \\ (2 \sqrt{x+3})'=2 (\sqrt{x+3})'=2 \frac{1}{2 \sqrt{x+3}} = \frac{1}{\sqrt{x+3}}$

93. freckles

$f''(x)=\frac{3 \cdot 2 \sqrt{x+3}-\frac{1 }{\sqrt{x+3}} \cdot (3x+6)}{4(x+3)}$

94. freckles

you multiply bottom and top by sqrt(x+3) to clear the compound fraction

95. anonymous

$(2\sqrt{X+3) (3) - }$

96. anonymous

what the heck...why didnt my answer show up

97. anonymous

okay ill do what you just asked

98. anonymous

so I got 6(sqrt x+3) * (sqrt x+3) - (3x-6)(sqrt x+3) all over 4(x+3)*(sqrt x+3)

99. anonymous

should I simplify more?

100. freckles

$f''(x)=\frac{6(x+3)-(3x+6)}{4 (x+3) \sqrt{x+3}}$

101. anonymous

yea, basicaly the same thing!

102. UsukiDoll

now we need critical points.

103. freckles

except you have a sqrt(x+3) next to the (3x+6)

104. anonymous

yea thanks for pointing that out!

105. freckles

anyways do you know where to go from here?

106. anonymous

so x= -4 is one of the critical points

107. freckles

find the POSSIBLE inflection points by finding where f''=0 f'' does not exist at x=-3 we already know at x=-3 we do not have an inflection point since it is an endpoint

108. freckles

$6(x+3)-3x-6=0 \\ 3x+12=0 \\ x=-4$ yeah sounds great

109. freckles

now to find if (-4,f(-4)) is an inflection points test the intervals to see if the concavity switches

110. anonymous

and for the undefined crtical point its going to x= -3?

111. anonymous

Do I also put x= -2 on the number line when testing out the points?

112. anonymous

|dw:1437454017504:dw|

113. anonymous

and then I test points in between those numbers with the second derivative to find where its CU and CD?

114. freckles

you have one possible inflection point so you only have two intervals to check

115. anonymous

so just stick with -3 and -2 ?

116. anonymous

okay

117. freckles

lol

118. freckles

wait sorry x=-4 wasn't on (-3,infty) which is the domain of f

119. freckles

we don't have an inflection point

120. freckles

x=-4 wasn't in the domain of the original function

121. anonymous

thats super odd

122. freckles

the function is either concave up or concave down

123. freckles

it doesn't have a combination of the two

124. anonymous

so the answer is like nothing

125. freckles

test any number in the domain of f that is test any number from (-3,inf) to see which it is

126. anonymous

I plugged in -3.5 to test it out between the intervals of (-3,-4)

127. freckles

put that number into f''

128. freckles

(-3,-4) doesn't make sense because -4 isn't greater than -3

129. anonymous

-4,-3

130. UsukiDoll

the bigger the negative number, the smaller its values is

131. freckles

but (-4,-3) isn't in the domain of f

132. anonymous

kk lemme plug in -2 then

133. freckles

you are forgetting the function has domain (-3,infty)

134. UsukiDoll

is it because at x =-3 the whole function is undefined?

135. freckles

you can choose a number from (-3,infty) to test the concavity of f by using f''

136. anonymous

so when I used -2 I got a negative number out which was -1.2

137. freckles

$f''(x)=\frac{6(x+3)-(3x+6)}{4 (x+3) \sqrt{x+3}}$ $f''(x)=\frac{3x+12}{4(x+3)\sqrt{x+3} } \\ f''(0)=\frac{3(0)+12}{4(0+3)\sqrt{0+3}} \\\ f''(0)=\frac{12}{4(3)\sqrt{3}}>0 \\ \text{ hmm... you got a negative number \let me see } \\ f''(-2)=\frac{3(-2)+12}{4(-2+3)\sqrt{-2+3}}=\frac{6}{4(1)\sqrt{1}}$ don't see how you got a negative number :(

138. anonymous

wow thats odd, I plugged the number into the first equation you put above but that shouldnt have given me a diff answer

139. freckles

you mean the unsimplfied version I have of f'' ?

140. anonymous

basicalllly

141. anonymous

okay i did it again and i got 6/4

142. anonymous

so this is how my number line looks like

143. freckles

how does it look?

144. anonymous

|dw:1437454970480:dw|

145. freckles

right because our function only exists to the right of and including x=-3

146. anonymous

sorry for the ratchetness of it, still getting used to it

147. freckles

what is all the + signs mean are you just saying it is concave up on (-3,infty)?

148. anonymous

ya

149. freckles

ok cool

150. anonymous

so is there also an inflection point at x=-3?

151. freckles

anyways I must leave you it is passed my bed time

152. freckles

no there is no inflection point

153. freckles

the concavity does not switch ever

154. nincompoop

laughing out loud

155. freckles

the function is concave up everywhere on the function's domain

156. anonymous

thanks a lot, you really are a great person! Bless yo soul dude

157. anonymous

LOL I really wish you were my calc teacher instead

158. freckles

it was no problem peace for now

159. anonymous

Peace freckles

160. nincompoop

ye a teacher that solves your own homework! that would be nice

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