anonymous
  • anonymous
What is the connection between critical numbers and relative extrema?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
matheducatorMcG
  • matheducatorMcG
Are you in calculus 1?
anonymous
  • anonymous
Yes
matheducatorMcG
  • 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Thank you so much! Is it possible for you to help me understand a few more problems?
matheducatorMcG
  • matheducatorMcG
keep on asking.
anonymous
  • anonymous
How do you find inflection points?
matheducatorMcG
  • matheducatorMcG
well, this occurs on curves that change concavity. this occurs when the second derivative, f"(x) is equal to zero.
anonymous
  • anonymous
Can you provide an example?
matheducatorMcG
  • matheducatorMcG
just a sec.
anonymous
  • anonymous
Okay
matheducatorMcG
  • 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
matheducatorMcG
  • matheducatorMcG
that's: possible inflection points are at
matheducatorMcG
  • 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.
anonymous
  • anonymous
What would it be in this case?
matheducatorMcG
  • 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
anonymous
  • anonymous
To list the inflection points, would I list them as: (-1,1) and (1,3)?
matheducatorMcG
  • matheducatorMcG
not quite. you must substitute the values of 0 and 2 into the original function to find the y value.
matheducatorMcG
  • matheducatorMcG
your points would be (0, ) and (2, )?
anonymous
  • anonymous
(0,0) and (2,0)
anonymous
  • anonymous
Correct?
matheducatorMcG
  • matheducatorMcG
ummm... not quite again. the original was\[f(x)=x^4-4x^3\] find f(0) and f(2).
matheducatorMcG
  • matheducatorMcG
(0,0) was right
anonymous
  • anonymous
(2, -16)
anonymous
  • anonymous
(2, -16)
matheducatorMcG
  • matheducatorMcG
ooohhh... so close. I get (2,-18).
anonymous
  • anonymous
I keep getting -16.
matheducatorMcG
  • matheducatorMcG
wait... I'm wrong... you're right
matheducatorMcG
  • matheducatorMcG
\[f(2)=(2)^4-4(2)^3\] Good Job!!
anonymous
  • 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.”
matheducatorMcG
  • 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.
anonymous
  • anonymous
So how would I correct the statement?
matheducatorMcG
  • matheducatorMcG
do we need to correct it to have the 'if ' statement or keep the 'then' statement?
anonymous
  • anonymous
It should be like... "So, the correct statement would be:...if...then..."
matheducatorMcG
  • 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.
matheducatorMcG
  • matheducatorMcG
the first I changed the 'then' statement. the second one i changed the 'if' statement.
anonymous
  • 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.”
anonymous
  • anonymous
This is for the second derivative.
matheducatorMcG
  • matheducatorMcG
false "If f''(c)<0 for all real numbers x, then f is concave down."
anonymous
  • anonymous
And why?
matheducatorMcG
  • matheducatorMcG
your 2nd derivative being negative for all values of x implies the whole graph is concave down
anonymous
  • 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)
matheducatorMcG
  • 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?
matheducatorMcG
  • matheducatorMcG
\[f(t)=t-4\sqrt{t+1}\] do you mean this?
anonymous
  • anonymous
Yes
matheducatorMcG
  • matheducatorMcG
do I have it right?
anonymous
  • anonymous
Yes
matheducatorMcG
  • matheducatorMcG
okay...just a sec
matheducatorMcG
  • matheducatorMcG
lost internet. typing ferociously now.
matheducatorMcG
  • matheducatorMcG
lets find the 1st derivative and set it equal to zero\[f(t)=t-4\sqrt{t+1}\]\[f '(t)=1-\frac{2}{\sqrt{t+1}}\]
matheducatorMcG
  • matheducatorMcG
\[0=1-\frac{2}{\sqrt{t+1}}\]
matheducatorMcG
  • matheducatorMcG
now add 1 to both sides, cross multiply and solve. let me know if you need the next step.
anonymous
  • anonymous
How would I cross multiply this?
matheducatorMcG
  • 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.
anonymous
  • anonymous
So I get: 2 / 1 times the root of (t+1)
matheducatorMcG
  • matheducatorMcG
\[\sqrt{t+1}=2\]
matheducatorMcG
  • matheducatorMcG
they become equal.
anonymous
  • anonymous
What's the next step?
matheducatorMcG
  • 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.
anonymous
  • anonymous
t=1
matheducatorMcG
  • matheducatorMcG
\[t+1=4\]
anonymous
  • anonymous
Oh yes! I forgot to square 2.
matheducatorMcG
  • matheducatorMcG
cool. good job fixing it.
matheducatorMcG
  • matheducatorMcG
so your extrema is at (3, ?)
anonymous
  • anonymous
I get: y = -2
matheducatorMcG
  • matheducatorMcG
hmmm... I get something else. \[f(3)=(3)-4\sqrt{(3)+1}\]
anonymous
  • anonymous
-1 to the root of 4 = -1 (2) = -2
matheducatorMcG
  • matheducatorMcG
order of ops. you have to evaluate the root first, then multiplication with the 4 then subtract from the 3
matheducatorMcG
  • matheducatorMcG
BEDMAS-brackets, exponents, division or multiplication, addition or subtraction.
anonymous
  • anonymous
(3, -5)
matheducatorMcG
  • 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.
matheducatorMcG
  • matheducatorMcG
so find f ' (2) and f ' (4) to see which pattern it takes on.
anonymous
  • anonymous
Plug 2 and 4 into the first derivative?
matheducatorMcG
  • 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]
anonymous
  • 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.
anonymous
  • anonymous
2 gives a negative answer and 4 gives a negative answer.
matheducatorMcG
  • matheducatorMcG
|dw:1341599783811:dw|
anonymous
  • anonymous
2 gives a negative answer and 4 gives a negative answer.
matheducatorMcG
  • matheducatorMcG
does 4 give us a negative in the 1st derivative? check again.
anonymous
  • anonymous
Wait nvm, it's positive. So it's negative to positive.
anonymous
  • anonymous
(3,-5) is a min.
matheducatorMcG
  • matheducatorMcG
yay! good job.
anonymous
  • anonymous
What about the maximum? There isn't one?
matheducatorMcG
  • matheducatorMcG
not unless you got more than one critical point. check the graph.
anonymous
  • anonymous
I did, I didn't think there was a max. I wanted to make sure I was right.
matheducatorMcG
  • matheducatorMcG
there is a vertical asymptote at x=-1. that would be the only other thing I would want to know
1 Attachment
anonymous
  • anonymous
okay
anonymous
  • anonymous
How do I find the limit for: limit as x approaches infinity 3x / (root of x^2+4)
matheducatorMcG
  • matheducatorMcG
do you all use L'Hopital's rule yet?
anonymous
  • anonymous
No
anonymous
  • anonymous
I have to find the limit analytically, without the L Hospital Rule.
matheducatorMcG
  • 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, \[\lim_{x \rightarrow \infty} \frac{3x}{\sqrt{x^2+4}}=3\]
anonymous
  • anonymous
Why wouldn't it be + and - 2?
anonymous
  • anonymous
OH! Wait, I got it.
anonymous
  • anonymous
Next problem... Find any vertical and horizontal asymptotes of the graph of the function. g(x) = 5x^2 / x^2+2
matheducatorMcG
  • 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..
anonymous
  • 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?
matheducatorMcG
  • 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.
anonymous
  • 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)
matheducatorMcG
  • matheducatorMcG
\[g(x)=x \sqrt{x+5}\] this is it?
anonymous
  • anonymous
yes
matheducatorMcG
  • matheducatorMcG
\[g'(x)=x \frac{1}{2\sqrt{x+5}}+\sqrt{x+5}\]
matheducatorMcG
  • matheducatorMcG
but we need the 2nd derivative, so...
matheducatorMcG
  • 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.
anonymous
  • anonymous
Okayy
matheducatorMcG
  • matheducatorMcG
combine the two fractions to get \[g"(x)=\frac{3x+20}{4(x+5)^{\frac{3}{2}}}\] then set equal to 0. \[0=\frac{3x+20}{4(x+5)^{\frac{3}{2}}}\]
matheducatorMcG
  • matheducatorMcG
the only way this is zero is if 3x+20 is equal to 0.
matheducatorMcG
  • matheducatorMcG
so inflection at x=? values on either side of the inflection value will tell you concave up or concave down.
anonymous
  • anonymous
3x+20 = 0 gives us -20/3
matheducatorMcG
  • matheducatorMcG
kew
anonymous
  • anonymous
So how would we find the inflection here?
matheducatorMcG
  • 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.
matheducatorMcG
  • matheducatorMcG
if positive, concave up. if negative, concave down.
anonymous
  • anonymous
I plugged it into my calculator and it keeps saying "domain error."
matheducatorMcG
  • matheducatorMcG
into the original function, right?
matheducatorMcG
  • 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
matheducatorMcG
  • matheducatorMcG
this means, there are no inflection points.
matheducatorMcG
  • matheducatorMcG
it has a min at -10/3 but is always concave up. check graph. when in doubt, graph it out.
matheducatorMcG
  • matheducatorMcG
your answer should be: there are no inflection points and there is a min at (-10/3, -4.303314829)
anonymous
  • anonymous
And for the concavity?
matheducatorMcG
  • matheducatorMcG
it is a global min, meaning there are no other min's. there are no global or relative max's.
matheducatorMcG
  • matheducatorMcG
if there is no inflection point and the graph has a min, then it is concave up.
matheducatorMcG
  • matheducatorMcG
*sorry, previous entry should read: if there is no inflection point and the graph has a global min, then it is concave up
anonymous
  • anonymous
Okay that's all?
matheducatorMcG
  • matheducatorMcG
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.
anonymous
  • 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.
matheducatorMcG
  • 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.
anonymous
  • anonymous
Okay, thank you so much once again!
anonymous
  • anonymous
Have a nice day.
matheducatorMcG
  • matheducatorMcG
you too.
matheducatorMcG
  • 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)

Looking for something else?

Not the answer you are looking for? Search for more explanations.