## hiralpatel121 Group Title What is the connection between critical numbers and relative extrema? 2 years ago 2 years ago

1. matheducatorMcG Group Title

Are you in calculus 1?

2. hiralpatel121 Group Title

Yes

3. matheducatorMcG Group Title

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. hiralpatel121 Group Title

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

5. matheducatorMcG Group Title

6. hiralpatel121 Group Title

How do you find inflection points?

7. matheducatorMcG Group Title

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

8. hiralpatel121 Group Title

Can you provide an example?

9. matheducatorMcG Group Title

just a sec.

10. hiralpatel121 Group Title

Okay

11. matheducatorMcG Group Title

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 Group Title

that's: possible inflection points are at

13. matheducatorMcG Group Title

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. hiralpatel121 Group Title

What would it be in this case?

15. matheducatorMcG Group Title

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. hiralpatel121 Group Title

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

17. matheducatorMcG Group Title

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

18. matheducatorMcG Group Title

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

19. hiralpatel121 Group Title

(0,0) and (2,0)

20. hiralpatel121 Group Title

Correct?

21. matheducatorMcG Group Title

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

22. matheducatorMcG Group Title

(0,0) was right

23. hiralpatel121 Group Title

(2, -16)

24. hiralpatel121 Group Title

(2, -16)

25. matheducatorMcG Group Title

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

26. hiralpatel121 Group Title

I keep getting -16.

27. matheducatorMcG Group Title

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

28. matheducatorMcG Group Title

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

29. hiralpatel121 Group Title

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 Group Title

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. hiralpatel121 Group Title

So how would I correct the statement?

32. matheducatorMcG Group Title

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

33. hiralpatel121 Group Title

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

34. matheducatorMcG Group Title

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 Group Title

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

36. hiralpatel121 Group Title

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. hiralpatel121 Group Title

This is for the second derivative.

38. matheducatorMcG Group Title

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

39. hiralpatel121 Group Title

And why?

40. matheducatorMcG Group Title

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

41. hiralpatel121 Group Title

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 Group Title

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 Group Title

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

44. hiralpatel121 Group Title

Yes

45. matheducatorMcG Group Title

do I have it right?

46. hiralpatel121 Group Title

Yes

47. matheducatorMcG Group Title

okay...just a sec

48. matheducatorMcG Group Title

lost internet. typing ferociously now.

49. matheducatorMcG Group Title

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 Group Title

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

51. matheducatorMcG Group Title

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

52. hiralpatel121 Group Title

How would I cross multiply this?

53. matheducatorMcG Group Title

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

54. hiralpatel121 Group Title

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

55. matheducatorMcG Group Title

$\sqrt{t+1}=2$

56. matheducatorMcG Group Title

they become equal.

57. hiralpatel121 Group Title

What's the next step?

58. matheducatorMcG Group Title

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. hiralpatel121 Group Title

t=1

60. matheducatorMcG Group Title

$t+1=4$

61. hiralpatel121 Group Title

Oh yes! I forgot to square 2.

62. matheducatorMcG Group Title

cool. good job fixing it.

63. matheducatorMcG Group Title

so your extrema is at (3, ?)

64. hiralpatel121 Group Title

I get: y = -2

65. matheducatorMcG Group Title

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

66. hiralpatel121 Group Title

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

67. matheducatorMcG Group Title

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

68. matheducatorMcG Group Title

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

69. hiralpatel121 Group Title

(3, -5)

70. matheducatorMcG Group Title

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 Group Title

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

72. hiralpatel121 Group Title

Plug 2 and 4 into the first derivative?

73. matheducatorMcG Group Title

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. hiralpatel121 Group Title

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. hiralpatel121 Group Title

76. matheducatorMcG Group Title

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77. hiralpatel121 Group Title

78. matheducatorMcG Group Title

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

79. hiralpatel121 Group Title

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

80. hiralpatel121 Group Title

(3,-5) is a min.

81. matheducatorMcG Group Title

yay! good job.

82. hiralpatel121 Group Title

What about the maximum? There isn't one?

83. matheducatorMcG Group Title

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

84. hiralpatel121 Group Title

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

85. matheducatorMcG Group Title

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

86. hiralpatel121 Group Title

okay

87. hiralpatel121 Group Title

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

88. matheducatorMcG Group Title

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

89. hiralpatel121 Group Title

No

90. hiralpatel121 Group Title

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

91. matheducatorMcG Group Title

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. hiralpatel121 Group Title

Why wouldn't it be + and - 2?

93. hiralpatel121 Group Title

OH! Wait, I got it.

94. hiralpatel121 Group Title

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

95. matheducatorMcG Group Title

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. hiralpatel121 Group Title

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 Group Title

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. hiralpatel121 Group Title

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 Group Title

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

100. hiralpatel121 Group Title

yes

101. matheducatorMcG Group Title

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

102. matheducatorMcG Group Title

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

103. matheducatorMcG Group Title

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

104. hiralpatel121 Group Title

Okayy

105. matheducatorMcG Group Title

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 Group Title

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

107. matheducatorMcG Group Title

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

108. hiralpatel121 Group Title

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

109. matheducatorMcG Group Title

kew

110. hiralpatel121 Group Title

So how would we find the inflection here?

111. matheducatorMcG Group Title

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 Group Title

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

113. hiralpatel121 Group Title

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

114. matheducatorMcG Group Title

into the original function, right?

115. matheducatorMcG Group Title

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 Group Title

this means, there are no inflection points.

117. matheducatorMcG Group Title

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

118. matheducatorMcG Group Title

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

119. hiralpatel121 Group Title

And for the concavity?

120. matheducatorMcG Group Title

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

121. matheducatorMcG Group Title

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

122. matheducatorMcG Group Title

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

123. hiralpatel121 Group Title

Okay that's all?

124. matheducatorMcG Group Title

125. hiralpatel121 Group Title

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 Group Title

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. hiralpatel121 Group Title

Okay, thank you so much once again!

128. hiralpatel121 Group Title

Have a nice day.

129. matheducatorMcG Group Title

you too.

130. matheducatorMcG Group Title

$h(t)=t-4\sqrt{t+1}$ $h'(t)=1-\frac{2}{\sqrt{t+1}}$ $0=1-\frac{2}{\sqrt{t+1}}$ $1=\frac{2}{\sqrt{t+1}}$ $2=\sqrt{t+1}$ $4=t+1$ $t=3$ $h''(t)=\frac{1}{(t+1)^{\frac{3}{2}}}$ $h''(3)=\frac{1}{8}$ 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)