Question

Need help on derivatives. attachment inside

Answer 1

which one

Answer 2

i need help on the concavity part on 11 and i also need help on number 21

Answer 3

f(x) = 6x - x^2

find f ' (x) = 0

Answer 4

for max/mins

and

f '' (x) = 0 to test for inflections

Answer 5

my favorite quote for concavity: *in a catchy little chant* "Concave up, Looks like a cup. Concave down, looks like a frown"

Answer 6

I'll let dan do the rest since he's more detailed already

Answer 7

It's my absolute favorite chant for calculus!

Answer 8

but this is better https://www.youtube.com/watch?v=uqwC41RDPyg

Answer 9

anyway this graph goes, it is differentiable.... to meee

Answer 10

Both of these probs, are just the power rule use to differentiate...

f(x) = 6x - x^2

f ' (x) = 6 - 2x

f ''(x) = -2


For relative Max/Min, set f'(x) to locate critical points to test for a change in increasing/decreasing

Answer 11

thrift shop is good too, but I digress. @OutSpoken find first and second derivatives then report back to us :)

Answer 12

f'(x)=6-2x and f''(x)=-2

Answer 13

f ' (x) = 0 = 6 - 2x
critical point - x = 3

Answer 14

test if f ' (x) is pos or negative on either side of the critical point x=3

Answer 15

wait

Answer 16

dont i have to do critical points for the second derivative

Answer 17

so f''(0)=-2 ?

Answer 18

yep, and it's negative right

Answer 19

ya

Answer 20

so, a negative second deriv implies what?

Answer 21

concave down

Answer 22

f ' (0) is not -2

f ' (x) = 6 - 2(0) = 6 (+)

Answer 23

^believe him, I didn't check your work.

Answer 24

i was gonna go with it till i just graphed the f(x). lol

Answer 25

but same method for f(0)=6

Answer 26

no i mean to find the critical points dont i set the second derivative to 0 ?

Answer 27

f''(0)=6

Answer 28

yeah, f ' (x) = 0 = 6 - 2x

x = 3

Answer 29

isnt the second derivative f;;(x)= -2 ?

Answer 30

no, cause it's a constant. Ie you have a car moving with constant acceleration a variable velocity and f(x) shows the position at any given time

Answer 31

also, we have the wrong function for 11, second deriv should have an x somewhere in general

Answer 32

There is no concavity switch, the second derivative is constant negative, always will be concave down

Answer 33

\[f(x)=x^3-6x^2+12x\]

Answer 34

according to the worksheet

Answer 35

i was doing 21, the whole time.

Answer 36

Im sorry can we start all over. Are we doing 11 or 21 first ?

Answer 37

lol, I was doing 11 hahahahah

Answer 38

f(x) = 6x - x^2

f ' (x) = 6 - 2x

f '' (x) = -2

------------------
f ' (x) = 0 when x = 3 (3 is the only critical point to test)

Answer 39

the first derivative f'(x) = 6 - 2x is positive for x is less than 3, and negative for x is larger than 3...

The function is increasing on the interval (-infinity, 3) and decreasing on interval (3,infinity)

Answer 40

the second derivative is a constant negative number, there is no inflection point, the function is always concave down. f ''(x) = -2

Answer 41

okay but the question is asking me to do the second derivative test so do i ignore the first derivative test in my final answer ?

Answer 42

The relative extrema - goes from increasing to decreasing - relative maximum

Answer 43

second derivative not applicable on this one, it is always concave down

Answer 44

The second derivative tells us concavity. The first derivative tells us slope changes

Answer 45

*It is concave down because the second derivative is always negative. The fact that it is negative is the second derivatives information. Since there are no critical points in the second derivative, we do not have to check anything else.*

Answer 46

setting f ' ' (x) = 0 will give you points of inflection on the graph. since f ' ' (x) = -2
0 = -2 ? there are no points of inflection for this function

Answer 47

The only extrema value is x = 3, the first derivative changes from + to - , the graph changes from increasing to decreasing slopes of the tangent. X=3 is a relative maximum.

Answer 48

if its changing from increasing to decresing doesnt that mean its a relative minimum?

Answer 49

and isnt the answer relative minimum (-infinity, 3) and relative max (3,infinity)

Answer 50

and where did the (3,9) come from ?