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What is the connection between critical numbers and relative extrema?
 one year ago
 one year ago
What is the connection between critical numbers and relative extrema?
 one year ago
 one year ago

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matheducatorMcGBest ResponseYou've already chosen the best response.2
Are you in calculus 1?
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
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.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
Thank you so much! Is it possible for you to help me understand a few more problems?
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
keep on asking.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
How do you find inflection points?
 one year ago

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

hiralpatel121Best ResponseYou've already chosen the best response.1
Can you provide an example?
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
determine 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
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
that's: possible inflection points are at
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
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.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
What would it be in this case?
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
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
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
To list the inflection points, would I list them as: (1,1) and (1,3)?
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
not quite. you must substitute the values of 0 and 2 into the original function to find the y value.
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
your points would be (0, ) and (2, )?
 one year ago

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

matheducatorMcGBest ResponseYou've already chosen the best response.2
(0,0) was right
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
ooohhh... so close. I get (2,18).
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
I keep getting 16.
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
wait... I'm wrong... you're right
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
\[f(2)=(2)^44(2)^3\] Good Job!!
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
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.”
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
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.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
So how would I correct the statement?
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
do we need to correct it to have the 'if ' statement or keep the 'then' statement?
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
It should be like... "So, the correct statement would be:...if...then..."
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
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.
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
the first I changed the 'then' statement. the second one i changed the 'if' statement.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
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.”
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
This is for the second derivative.
 one year ago

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

matheducatorMcGBest ResponseYou've already chosen the best response.2
your 2nd derivative being negative for all values of x implies the whole graph is concave down
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
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)
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
f(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?
 one year ago

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

matheducatorMcGBest ResponseYou've already chosen the best response.2
do I have it right?
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
okay...just a sec
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
lost internet. typing ferociously now.
 one year ago

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

matheducatorMcGBest ResponseYou've already chosen the best response.2
\[0=1\frac{2}{\sqrt{t+1}}\]
 one year ago

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

hiralpatel121Best ResponseYou've already chosen the best response.1
How would I cross multiply this?
 one year ago

matheducatorMcGBest 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.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
So I get: 2 / 1 times the root of (t+1)
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
\[\sqrt{t+1}=2\]
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
they become equal.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
What's the next step?
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
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.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
Oh yes! I forgot to square 2.
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
cool. good job fixing it.
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
so your extrema is at (3, ?)
 one year ago

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

hiralpatel121Best ResponseYou've already chosen the best response.1
1 to the root of 4 = 1 (2) = 2
 one year ago

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

matheducatorMcGBest ResponseYou've already chosen the best response.2
BEDMASbrackets, exponents, division or multiplication, addition or subtraction.
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
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.
 one year ago

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

hiralpatel121Best ResponseYou've already chosen the best response.1
Plug 2 and 4 into the first derivative?
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
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]
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
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.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
2 gives a negative answer and 4 gives a negative answer.
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
dw:1341599783811:dw
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
2 gives a negative answer and 4 gives a negative answer.
 one year ago

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

hiralpatel121Best ResponseYou've already chosen the best response.1
Wait nvm, it's positive. So it's negative to positive.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
(3,5) is a min.
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
yay! good job.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
What about the maximum? There isn't one?
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
not unless you got more than one critical point. check the graph.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
I did, I didn't think there was a max. I wanted to make sure I was right.
 one year ago

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

hiralpatel121Best ResponseYou've already chosen the best response.1
How do I find the limit for: limit as x approaches infinity 3x / (root of x^2+4)
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
do you all use L'Hopital's rule yet?
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
I have to find the limit analytically, without the L Hospital Rule.
 one year ago

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

hiralpatel121Best ResponseYou've already chosen the best response.1
Why wouldn't it be + and  2?
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
OH! Wait, I got it.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
Next problem... Find any vertical and horizontal asymptotes of the graph of the function. g(x) = 5x^2 / x^2+2
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
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..
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
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?
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
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.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
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)
 one year ago

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

matheducatorMcGBest ResponseYou've already chosen the best response.2
\[g'(x)=x \frac{1}{2\sqrt{x+5}}+\sqrt{x+5}\]
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
but we need the 2nd derivative, so...
 one year ago

matheducatorMcGBest 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.
 one year ago

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

matheducatorMcGBest ResponseYou've already chosen the best response.2
the only way this is zero is if 3x+20 is equal to 0.
 one year ago

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

hiralpatel121Best ResponseYou've already chosen the best response.1
3x+20 = 0 gives us 20/3
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
So how would we find the inflection here?
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
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.
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
if positive, concave up. if negative, concave down.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
I plugged it into my calculator and it keeps saying "domain error."
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
into the original function, right?
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
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
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
this means, there are no inflection points.
 one year ago

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

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

hiralpatel121Best ResponseYou've already chosen the best response.1
And for the concavity?
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
it is a global min, meaning there are no other min's. there are no global or relative max's.
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
if there is no inflection point and the graph has a min, then it is concave up.
 one year ago

matheducatorMcGBest 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
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
Okay that's all?
 one year ago

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

hiralpatel121Best ResponseYou've already chosen the best response.1
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.
 one year ago

matheducatorMcGBest ResponseYou've already chosen the best response.2
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.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
Okay, thank you so much once again!
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
Have a nice day.
 one year ago

matheducatorMcGBest 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)
 one year ago
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