Question

What is the connection between critical numbers and relative extrema?

Answer 1

Are you in calculus 1?

Answer 2

Yes

Answer 3

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.

Answer 4

Thank you so much! Is it possible for you to help me understand a few more problems?

Answer 5

keep on asking.

Answer 6

How do you find inflection points?

Answer 7

well, this occurs on curves that change concavity. this occurs when the second derivative, f"(x) is equal to zero.

Answer 8

Can you provide an example?

Answer 9

just a sec.

Answer 10

Okay

Answer 11

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

Answer 12

that's: possible inflection points are at

Answer 13

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.

Answer 14

What would it be in this case?

Answer 15

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

Answer 16

To list the inflection points, would I list them as: (-1,1) and (1,3)?

Answer 17

not quite. you must substitute the values of 0 and 2 into the original function to find the y value.

Answer 18

your points would be (0, ) and (2, )?

Answer 19

(0,0) and (2,0)

Answer 20

Correct?

Answer 21

ummm... not quite again.
the original was\[f(x)=x^4-4x^3\]
find f(0) and f(2).

Answer 22

(0,0) was right

Answer 23

(2, -16)

Answer 24

(2, -16)

Answer 25

ooohhh... so close.
I get (2,-18).

Answer 26

I keep getting -16.

Answer 27

wait... I'm wrong... you're right

Answer 28

\[f(2)=(2)^4-4(2)^3\]
Good Job!!

Answer 29

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.”

Answer 30

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.

Answer 31

So how would I correct the statement?

Answer 32

do we need to correct it to have the 'if ' statement or keep the 'then' statement?

Answer 33

It should be like... "So, the correct statement would be:...if...then..."

Answer 34

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.

Answer 35

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

Answer 36

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.”

Answer 37

This is for the second derivative.

Answer 38

false
"If f''(c)<0 for all real numbers x, then f is concave down."

Answer 39

And why?

Answer 40

your 2nd derivative being negative for all values of x implies the whole graph is concave down

Answer 41

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)

Answer 42

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?

Answer 43

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

Answer 44

Yes

Answer 45

do I have it right?

Answer 46

Yes

Answer 47

okay...just a sec

Answer 48

lost internet. typing ferociously now.

Answer 49

lets find the 1st derivative and set it equal to zero\[f(t)=t-4\sqrt{t+1}\]\[f '(t)=1-\frac{2}{\sqrt{t+1}}\]

Answer 50

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

Answer 51

now add 1 to both sides, cross multiply and solve. let me know if you need the next step.

Answer 52

How would I cross multiply this?

Answer 53

\[1=\frac{2}{\sqrt{t+1}}\]
put the 1 over 1 and multiply, remembering that the products you get are equal to each other.

Answer 54

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

Answer 55

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

Answer 56

they become equal.

Answer 57

What's the next step?

Answer 58

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.

Answer 59

t=1

Answer 60

\[t+1=4\]

Answer 61

Oh yes! I forgot to square 2.

Answer 62

cool. good job fixing it.

Answer 63

so your extrema is at (3, ?)

Answer 64

I get: y = -2

Answer 65

hmmm... I get something else.
\[f(3)=(3)-4\sqrt{(3)+1}\]

Answer 66

-1 to the root of 4 = -1 (2) = -2

Answer 67

order of ops. you have to evaluate the root first, then multiplication with the 4 then subtract from the 3

Answer 68

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

Answer 69

(3, -5)

Answer 70

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.

Answer 71

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

Answer 72

Plug 2 and 4 into the first derivative?

Answer 73

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]

Answer 74

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.

Answer 75

2 gives a negative answer and 4 gives a negative answer.