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

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matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2Are you in calculus 1?

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2critical 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thank you so much! Is it possible for you to help me understand a few more problems?

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2keep on asking.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How do you find inflection points?

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2well, this occurs on curves that change concavity. this occurs when the second derivative, f"(x) is equal to zero.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Can you provide an example?

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2determine points of inflection for \[f(x)=x^44x^3\] find second derivative \[f''(x)=12x^224x=12x(x2)\] 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

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2that's: possible inflection points are at

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2if 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What would it be in this case?

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2f''(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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0To list the inflection points, would I list them as: (1,1) and (1,3)?

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2not quite. you must substitute the values of 0 and 2 into the original function to find the y value.

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2your points would be (0, ) and (2, )?

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2ummm... not quite again. the original was\[f(x)=x^44x^3\] find f(0) and f(2).

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2(0,0) was right

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2ooohhh... so close. I get (2,18).

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2wait... I'm wrong... you're right

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2\[f(2)=(2)^44(2)^3\] Good Job!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, 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.”

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2sorry 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So how would I correct the statement?

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2do we need to correct it to have the 'if ' statement or keep the 'then' statement?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It should be like... "So, the correct statement would be:...if...then..."

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2i'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.

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2the first I changed the 'then' statement. the second one i changed the 'if' statement.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay 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.”

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is for the second derivative.

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2false "If f''(c)<0 for all real numbers x, then f is concave down."

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2your 2nd derivative being negative for all values of x implies the whole graph is concave down

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay. I don’t know how to find the relative extrema for the following function. f(t) = t – 4 to the root of ( t+1)

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2f(t)=t4 to the root of (t+1) to the root? does that mean multiplied by the root or to the power of the root?

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2\[f(t)=t4\sqrt{t+1}\] do you mean this?

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2do I have it right?

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2okay...just a sec

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2lost internet. typing ferociously now.

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2lets find the 1st derivative and set it equal to zero\[f(t)=t4\sqrt{t+1}\]\[f '(t)=1\frac{2}{\sqrt{t+1}}\]

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2\[0=1\frac{2}{\sqrt{t+1}}\]

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2now add 1 to both sides, cross multiply and solve. let me know if you need the next step.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How would I cross multiply this?

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2\[1=\frac{2}{\sqrt{t+1}}\] put the 1 over 1 and multiply, remembering that the products you get are equal to each other.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So I get: 2 / 1 times the root of (t+1)

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2\[\sqrt{t+1}=2\]

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2they become equal.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What's the next step?

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2now 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh yes! I forgot to square 2.

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2cool. good job fixing it.

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2so your extrema is at (3, ?)

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2hmmm... I get something else. \[f(3)=(3)4\sqrt{(3)+1}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.01 to the root of 4 = 1 (2) = 2

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2order of ops. you have to evaluate the root first, then multiplication with the 4 then subtract from the 3

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2BEDMASbrackets, exponents, division or multiplication, addition or subtraction.

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2YAY! 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.

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2so find f ' (2) and f ' (4) to see which pattern it takes on.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Plug 2 and 4 into the first derivative?

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2yup. 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]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0When 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.02 gives a negative answer and 4 gives a negative answer.

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2dw:1341599783811:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.02 gives a negative answer and 4 gives a negative answer.

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2does 4 give us a negative in the 1st derivative? check again.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wait nvm, it's positive. So it's negative to positive.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What about the maximum? There isn't one?

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2not unless you got more than one critical point. check the graph.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I did, I didn't think there was a max. I wanted to make sure I was right.

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2there is a vertical asymptote at x=1. that would be the only other thing I would want to know

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How do I find the limit for: limit as x approaches infinity 3x / (root of x^2+4)

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2do you all use L'Hopital's rule yet?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have to find the limit analytically, without the L Hospital Rule.

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2okay... 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, \[\lim_{x \rightarrow \infty} \frac{3x}{\sqrt{x^2+4}}=3\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Why wouldn't it be + and  2?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Next problem... Find any vertical and horizontal asymptotes of the graph of the function. g(x) = 5x^2 / x^2+2

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2horizontal 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..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0One question.. There are times when we just set the top and bottom equal to 0 and find asymptotes. Does this occur all the time?

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2Rational 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, 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)

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2\[g(x)=x \sqrt{x+5}\] this is it?

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2\[g'(x)=x \frac{1}{2\sqrt{x+5}}+\sqrt{x+5}\]

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2but we need the 2nd derivative, so...

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2\[g"(x)=\frac{x}{4(x+5)^{\frac{3}{2}}}+\frac{1}{\sqrt{x+5}} \] now set equal to zero and solve. whooo.... give me a sec here.

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2combine the two fractions to get \[g"(x)=\frac{3x+20}{4(x+5)^{\frac{3}{2}}}\] then set equal to 0. \[0=\frac{3x+20}{4(x+5)^{\frac{3}{2}}}\]

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2the only way this is zero is if 3x+20 is equal to 0.

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2so inflection at x=? values on either side of the inflection value will tell you concave up or concave down.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.03x+20 = 0 gives us 20/3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So how would we find the inflection here?

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2x=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.

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2if positive, concave up. if negative, concave down.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I plugged it into my calculator and it keeps saying "domain error."

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2into the original function, right?

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2oh 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

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2this means, there are no inflection points.

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2it has a min at 10/3 but is always concave up. check graph. when in doubt, graph it out.

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2your answer should be: there are no inflection points and there is a min at (10/3, 4.303314829)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And for the concavity?

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2it is a global min, meaning there are no other min's. there are no global or relative max's.

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2if there is no inflection point and the graph has a min, then it is concave up.

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2*sorry, previous entry should read: if there is no inflection point and the graph has a global min, then it is concave up

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2http://www.wolframalpha.com/ can help you immensely if your looking to get quick answers. more explanation may be needed but its a great start.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thank 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.

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2I 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, thank you so much once again!

matheducatorMcG
 4 years ago
Best ResponseYou've already chosen the best response.2\[h(t)=t4\sqrt{t+1}\] \[h'(t)=1\frac{2}{\sqrt{t+1}}\] \[0=1\frac{2}{\sqrt{t+1}}\] \[1=\frac{2}{\sqrt{t+1}}\] \[2=\sqrt{t+1}\] \[4=t+1\] \[t=3\] \[h''(t)=\frac{1}{(t+1)^{\frac{3}{2}}}\] \[h''(3)=\frac{1}{8}\] 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)
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