What is the connection between critical numbers and relative extrema?

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What is the connection between critical numbers and relative extrema?

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Are you in calculus 1?
Yes
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.

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Thank you so much! Is it possible for you to help me understand a few more problems?
keep on asking.
How do you find inflection points?
well, this occurs on curves that change concavity. this occurs when the second derivative, f"(x) is equal to zero.
Can you provide an example?
just a sec.
Okay
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
that's: possible inflection points are at
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.
What would it be in this case?
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
To list the inflection points, would I list them as: (-1,1) and (1,3)?
not quite. you must substitute the values of 0 and 2 into the original function to find the y value.
your points would be (0, ) and (2, )?
(0,0) and (2,0)
Correct?
ummm... not quite again. the original was\[f(x)=x^4-4x^3\] find f(0) and f(2).
(0,0) was right
(2, -16)
(2, -16)
ooohhh... so close. I get (2,-18).
I keep getting -16.
wait... I'm wrong... you're right
\[f(2)=(2)^4-4(2)^3\] Good Job!!
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.”
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.
So how would I correct the statement?
do we need to correct it to have the 'if ' statement or keep the 'then' statement?
It should be like... "So, the correct statement would be:...if...then..."
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.
the first I changed the 'then' statement. the second one i changed the 'if' statement.
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.”
This is for the second derivative.
false "If f''(c)<0 for all real numbers x, then f is concave down."
And why?
your 2nd derivative being negative for all values of x implies the whole graph is concave down
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)
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?
\[f(t)=t-4\sqrt{t+1}\] do you mean this?
Yes
do I have it right?
Yes
okay...just a sec
lost internet. typing ferociously now.
lets find the 1st derivative and set it equal to zero\[f(t)=t-4\sqrt{t+1}\]\[f '(t)=1-\frac{2}{\sqrt{t+1}}\]
\[0=1-\frac{2}{\sqrt{t+1}}\]
now add 1 to both sides, cross multiply and solve. let me know if you need the next step.
How would I cross multiply this?
\[1=\frac{2}{\sqrt{t+1}}\] put the 1 over 1 and multiply, remembering that the products you get are equal to each other.
So I get: 2 / 1 times the root of (t+1)
\[\sqrt{t+1}=2\]
they become equal.
What's the next step?
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.
t=1
\[t+1=4\]
Oh yes! I forgot to square 2.
cool. good job fixing it.
so your extrema is at (3, ?)
I get: y = -2
hmmm... I get something else. \[f(3)=(3)-4\sqrt{(3)+1}\]
-1 to the root of 4 = -1 (2) = -2
order of ops. you have to evaluate the root first, then multiplication with the 4 then subtract from the 3
BEDMAS-brackets, exponents, division or multiplication, addition or subtraction.
(3, -5)
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.
so find f ' (2) and f ' (4) to see which pattern it takes on.
Plug 2 and 4 into the first derivative?
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]
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.
2 gives a negative answer and 4 gives a negative answer.
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2 gives a negative answer and 4 gives a negative answer.
does 4 give us a negative in the 1st derivative? check again.
Wait nvm, it's positive. So it's negative to positive.
(3,-5) is a min.
yay! good job.
What about the maximum? There isn't one?
not unless you got more than one critical point. check the graph.
I did, I didn't think there was a max. I wanted to make sure I was right.
there is a vertical asymptote at x=-1. that would be the only other thing I would want to know
1 Attachment
okay
How do I find the limit for: limit as x approaches infinity 3x / (root of x^2+4)
do you all use L'Hopital's rule yet?
No
I have to find the limit analytically, without the L Hospital Rule.
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, \[\lim_{x \rightarrow \infty} \frac{3x}{\sqrt{x^2+4}}=3\]
Why wouldn't it be + and - 2?
OH! Wait, I got it.
Next problem... Find any vertical and horizontal asymptotes of the graph of the function. g(x) = 5x^2 / x^2+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 question.. There are times when we just set the top and bottom equal to 0 and find asymptotes. Does this occur all the time?
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.
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)
\[g(x)=x \sqrt{x+5}\] this is it?
yes
\[g'(x)=x \frac{1}{2\sqrt{x+5}}+\sqrt{x+5}\]
but we need the 2nd derivative, so...
\[g"(x)=-\frac{x}{4(x+5)^{\frac{3}{2}}}+\frac{1}{\sqrt{x+5}} \] now set equal to zero and solve. whooo.... give me a sec here.
Okayy
combine the two fractions to get \[g"(x)=\frac{3x+20}{4(x+5)^{\frac{3}{2}}}\] then set equal to 0. \[0=\frac{3x+20}{4(x+5)^{\frac{3}{2}}}\]
the only way this is zero is if 3x+20 is equal to 0.
so inflection at x=? values on either side of the inflection value will tell you concave up or concave down.
3x+20 = 0 gives us -20/3
kew
So how would we find the inflection here?
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.
if positive, concave up. if negative, concave down.
I plugged it into my calculator and it keeps saying "domain error."
into the original function, right?
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
this means, there are no inflection points.
it has a min at -10/3 but is always concave up. check graph. when in doubt, graph it out.
your answer should be: there are no inflection points and there is a min at (-10/3, -4.303314829)
And for the concavity?
it is a global min, meaning there are no other min's. there are no global or relative max's.
if there is no inflection point and the graph has a min, then it is concave up.
*sorry, previous entry should read: if there is no inflection point and the graph has a global min, then it is concave up
Okay that's all?
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.
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.
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.
Okay, thank you so much once again!
Have a nice day.
you too.
\[h(t)=t-4\sqrt{t+1}\] \[h'(t)=1-\frac{2}{\sqrt{t+1}}\] \[0=1-\frac{2}{\sqrt{t+1}}\] \[1=\frac{2}{\sqrt{t+1}}\] \[2=\sqrt{t+1}\] \[4=t+1\] \[t=3\] \[h''(t)=\frac{1}{(t+1)^{\frac{3}{2}}}\] \[h''(3)=\frac{1}{8}\] 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)

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