## anonymous 4 years ago What is the connection between critical numbers and relative extrema?

1. matheducatorMcG

Are you in calculus 1?

2. anonymous

Yes

3. matheducatorMcG

critical numbers can tell you where relative extrema occur on the graph. we use derivatives to find these critical numbers which in turn allows us to better sketch curves.

4. anonymous

Thank you so much! Is it possible for you to help me understand a few more problems?

5. matheducatorMcG

6. anonymous

How do you find inflection points?

7. matheducatorMcG

well, this occurs on curves that change concavity. this occurs when the second derivative, f"(x) is equal to zero.

8. anonymous

Can you provide an example?

9. matheducatorMcG

just a sec.

10. anonymous

Okay

11. matheducatorMcG

determine points of inflection for $f(x)=x^4-4x^3$ find second derivative $f''(x)=12x^2-24x=12x(x-2)$ set equal to zero and solve. possible points are at $x=0$and $x=2$ Now test values around 0 and 2, like -1 and 1 and 3

12. matheducatorMcG

that's: possible inflection points are at

13. matheducatorMcG

if f''(x) >0 in the interval, concave up. if f''(x) <0 in the interval, concave down. if the values go concave up then concave down, you've got an inflection point.

14. anonymous

What would it be in this case?

15. matheducatorMcG

f''(-1)>0(concave up) and f''(1)<0(concave down) and f''(3)>0(concave up). Therefore, we can conclude that both are inflection points because we go concave up->concave down->concave up

16. anonymous

To list the inflection points, would I list them as: (-1,1) and (1,3)?

17. matheducatorMcG

not quite. you must substitute the values of 0 and 2 into the original function to find the y value.

18. matheducatorMcG

your points would be (0, ) and (2, )?

19. anonymous

(0,0) and (2,0)

20. anonymous

Correct?

21. matheducatorMcG

ummm... not quite again. the original was$f(x)=x^4-4x^3$ find f(0) and f(2).

22. matheducatorMcG

(0,0) was right

23. anonymous

(2, -16)

24. anonymous

(2, -16)

25. matheducatorMcG

ooohhh... so close. I get (2,-18).

26. anonymous

I keep getting -16.

27. matheducatorMcG

wait... I'm wrong... you're right

28. matheducatorMcG

$f(2)=(2)^4-4(2)^3$ Good Job!!

29. anonymous

Okay, thank you! The next thing I need help with is: Is the following true or false? If false, explain why and correct the statement. “If f ‘ (c)<0, then f is concave down at x=c.”

30. matheducatorMcG

sorry hold on. I miss read the question real quick. f'(c)<0 just means the slope of the curve at that point is negative. a parabola has some c where f'(c)<0 but the entire graph is always concave up. false.

31. anonymous

So how would I correct the statement?

32. matheducatorMcG

do we need to correct it to have the 'if ' statement or keep the 'then' statement?

33. anonymous

It should be like... "So, the correct statement would be:...if...then..."

34. matheducatorMcG

i'll give both "If f '(c)<0, then f is decreasing at x=c." "If f ''(c)<0, then f is concave down." i think those suffice.

35. matheducatorMcG

the first I changed the 'then' statement. the second one i changed the 'if' statement.

36. anonymous

Okay thank you. What about.. Is the following true or false? If false, explain why and correct the statement. “If f ‘’(c)<0, for all real numbers x, then f decreases without a bound.”

37. anonymous

This is for the second derivative.

38. matheducatorMcG

false "If f''(c)<0 for all real numbers x, then f is concave down."

39. anonymous

And why?

40. matheducatorMcG

your 2nd derivative being negative for all values of x implies the whole graph is concave down

41. anonymous

Okay. I don’t know how to find the relative extrema for the following function. f(t) = t – 4 to the root of ( t+1)

42. matheducatorMcG

f(t)=t-4 to the root of (t+1) to the root? does that mean multiplied by the root or to the power of the root?

43. matheducatorMcG

$f(t)=t-4\sqrt{t+1}$ do you mean this?

44. anonymous

Yes

45. matheducatorMcG

do I have it right?

46. anonymous

Yes

47. matheducatorMcG

okay...just a sec

48. matheducatorMcG

lost internet. typing ferociously now.

49. matheducatorMcG

lets find the 1st derivative and set it equal to zero$f(t)=t-4\sqrt{t+1}$$f '(t)=1-\frac{2}{\sqrt{t+1}}$

50. matheducatorMcG

$0=1-\frac{2}{\sqrt{t+1}}$

51. matheducatorMcG

now add 1 to both sides, cross multiply and solve. let me know if you need the next step.

52. anonymous

How would I cross multiply this?

53. matheducatorMcG

$1=\frac{2}{\sqrt{t+1}}$ put the 1 over 1 and multiply, remembering that the products you get are equal to each other.

54. anonymous

So I get: 2 / 1 times the root of (t+1)

55. matheducatorMcG

$\sqrt{t+1}=2$

56. matheducatorMcG

they become equal.

57. anonymous

What's the next step?

58. matheducatorMcG

now square both sides and solve for x. this tells you the x value where there is extrema. now substitute x into f(t) to find your y.

59. anonymous

t=1

60. matheducatorMcG

$t+1=4$

61. anonymous

Oh yes! I forgot to square 2.

62. matheducatorMcG

cool. good job fixing it.

63. matheducatorMcG

so your extrema is at (3, ?)

64. anonymous

I get: y = -2

65. matheducatorMcG

hmmm... I get something else. $f(3)=(3)-4\sqrt{(3)+1}$

66. anonymous

-1 to the root of 4 = -1 (2) = -2

67. matheducatorMcG

order of ops. you have to evaluate the root first, then multiplication with the 4 then subtract from the 3

68. matheducatorMcG

BEDMAS-brackets, exponents, division or multiplication, addition or subtraction.

69. anonymous

(3, -5)

70. matheducatorMcG

YAY! now, is this a min or a max? we can check by substituting values on either side of 3 into f '(t) to see if it goes increasing to decreasing OR decreasing to increasing. that will tell us if it is a min or a max.

71. matheducatorMcG

so find f ' (2) and f ' (4) to see which pattern it takes on.

72. anonymous

Plug 2 and 4 into the first derivative?

73. matheducatorMcG

yup. the 1st derivative is the slope of the curve. if it goes decreasing to increasing, its a min. if it goes increasing to decreasing, it is a max. you could also look at the 2nd derivative to see if it is positive or negative at that point, but this doesn't always work. it works here though. ;0]

74. anonymous

When I plugged 2 in, I got 1 - 2 times root of 3. When I plugged 4 in, I got 1 - 2 times root of 5.

75. anonymous

76. matheducatorMcG

|dw:1341599783811:dw|

77. anonymous

78. matheducatorMcG

does 4 give us a negative in the 1st derivative? check again.

79. anonymous

Wait nvm, it's positive. So it's negative to positive.

80. anonymous

(3,-5) is a min.

81. matheducatorMcG

yay! good job.

82. anonymous

What about the maximum? There isn't one?

83. matheducatorMcG

not unless you got more than one critical point. check the graph.

84. anonymous

I did, I didn't think there was a max. I wanted to make sure I was right.

85. matheducatorMcG

there is a vertical asymptote at x=-1. that would be the only other thing I would want to know

86. anonymous

okay

87. anonymous

How do I find the limit for: limit as x approaches infinity 3x / (root of x^2+4)

88. matheducatorMcG

do you all use L'Hopital's rule yet?

89. anonymous

No

90. anonymous

I have to find the limit analytically, without the L Hospital Rule.

91. matheducatorMcG

okay... how about the fact that were going off to infinity so, $\sqrt{x^2} \approx \sqrt{x^2+4}$ as x-> infinity, therefore, we can evaluate $\lim_{x \rightarrow \infty} \frac{3x}{\sqrt{x^2+4}}$ as $\lim_{x \rightarrow \infty} \frac{3x}{\sqrt{x^2}}$ which algebraically simplifies to be $\lim_{x \rightarrow \infty} \frac{3x}{x}$ which again simplifies to 3. so, $\lim_{x \rightarrow \infty} \frac{3x}{\sqrt{x^2+4}}=3$

92. anonymous

Why wouldn't it be + and - 2?

93. anonymous

OH! Wait, I got it.

94. anonymous

Next problem... Find any vertical and horizontal asymptotes of the graph of the function. g(x) = 5x^2 / x^2+2

95. matheducatorMcG

horizontal asymptote is what value you approach as x->infinity, mainly 5. vertical asymptotes are what makes the denominator zero, so none here. asymptotes are based off limits as x approaches infinity for horizontal or as x approaches some value a for vertical asymptotes..

96. anonymous

One question.. There are times when we just set the top and bottom equal to 0 and find asymptotes. Does this occur all the time?

97. matheducatorMcG

Rational functions are the graphs that have the most asymptotes. But be careful. Read on. With rational functions the denominator will tell you where vertical asymptotes are located. setting the numerator equal to zero will tell you where it crosses the x axis. finding horizontal asymptotes are all about the limit of the graph as x approaches infinity.

98. anonymous

Okay, one last question I need help with. What would be the points of inflection and concavity of the following graph of the function? g(x) = x to the root of (x+5)

99. matheducatorMcG

$g(x)=x \sqrt{x+5}$ this is it?

100. anonymous

yes

101. matheducatorMcG

$g'(x)=x \frac{1}{2\sqrt{x+5}}+\sqrt{x+5}$

102. matheducatorMcG

but we need the 2nd derivative, so...

103. matheducatorMcG

$g"(x)=-\frac{x}{4(x+5)^{\frac{3}{2}}}+\frac{1}{\sqrt{x+5}}$ now set equal to zero and solve. whooo.... give me a sec here.

104. anonymous

Okayy

105. matheducatorMcG

combine the two fractions to get $g"(x)=\frac{3x+20}{4(x+5)^{\frac{3}{2}}}$ then set equal to 0. $0=\frac{3x+20}{4(x+5)^{\frac{3}{2}}}$

106. matheducatorMcG

the only way this is zero is if 3x+20 is equal to 0.

107. matheducatorMcG

so inflection at x=? values on either side of the inflection value will tell you concave up or concave down.

108. anonymous

3x+20 = 0 gives us -20/3

109. matheducatorMcG

kew

110. anonymous

So how would we find the inflection here?

111. matheducatorMcG

x=-20/3 gets substituted into the original function to find the y value of the point. then test values on either side of the inflection point in the 2nd derivative to see if positive or negative.

112. matheducatorMcG

if positive, concave up. if negative, concave down.

113. anonymous

I plugged it into my calculator and it keeps saying "domain error."

114. matheducatorMcG

into the original function, right?

115. matheducatorMcG

oh wait... the calculator is right. -20/3 would make the inside of the root negative, which is no no in the real world. hmm... contemplating

116. matheducatorMcG

this means, there are no inflection points.

117. matheducatorMcG

it has a min at -10/3 but is always concave up. check graph. when in doubt, graph it out.

118. matheducatorMcG

your answer should be: there are no inflection points and there is a min at (-10/3, -4.303314829)

119. anonymous

And for the concavity?

120. matheducatorMcG

it is a global min, meaning there are no other min's. there are no global or relative max's.

121. matheducatorMcG

if there is no inflection point and the graph has a min, then it is concave up.

122. matheducatorMcG

*sorry, previous entry should read: if there is no inflection point and the graph has a global min, then it is concave up

123. anonymous

Okay that's all?

124. matheducatorMcG

125. anonymous

Thank you so much for your help! I have a test next week and I might come across a few problems like this. Will you be online next week? I'd love if you could help me solve a few more problems so I have more to study from.

126. matheducatorMcG

I should be as this website is my new internet addiction. I am teaching a small class this summer but I love answering math questions, so I'll be around. become a fan of mine and send me a message if you get on and see me.

127. anonymous

Okay, thank you so much once again!

128. anonymous

Have a nice day.

129. matheducatorMcG

you too.

130. matheducatorMcG

$h(t)=t-4\sqrt{t+1}$ $h'(t)=1-\frac{2}{\sqrt{t+1}}$ $0=1-\frac{2}{\sqrt{t+1}}$ $1=\frac{2}{\sqrt{t+1}}$ $2=\sqrt{t+1}$ $4=t+1$ $t=3$ $h''(t)=\frac{1}{(t+1)^{\frac{3}{2}}}$ $h''(3)=\frac{1}{8}$ which is positive, so by the 2nd derivative test, this is a min. there is no relative max. the min occurs at $h(3)=-5$ (3,-5)