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Are you in calculus 1?

Yes

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?

Can you provide an example?

just a sec.

Okay

that's: possible inflection points are at

What would it be in this case?

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!!

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..."

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

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

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

Yes

do I have it right?

Yes

okay...just a sec

lost internet. typing ferociously now.

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

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?

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

\[\sqrt{t+1}=2\]

they become equal.

What's the next step?

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

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

(3, -5)

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

Plug 2 and 4 into the first derivative?

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.

|dw:1341599783811:dw|

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

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.

Why wouldn't it be + and - 2?

OH! Wait, I got it.

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

yes

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

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

Okayy

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

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

kew

So how would we find the inflection here?

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?

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.

Okay that's all?

Okay, thank you so much once again!

Have a nice day.

you too.