Question

Identify a function f that has the given characteristics. f(–5) = f(3) = 0; f '(-1) = 0, f '(x) < 0 for x < -1; f '(x) > 0 for x > -1 f(x) =

Answer 1

this is a different question

Answer 2

what do you mean?

Answer 3

can you help with any of these?:)

Answer 4

sure lets do the screenshot one

Answer 5

great!!!!!

Answer 6

point is \((2,8)\) slope is \(f'(2)\) point slope formula will give the equation of the line

Answer 7

for the fist one you posted
you know what \(f'(x)\) is in this case?

Answer 8

i posted a screen shot of that one as well just now.

Answer 9

ok lets do them one at a time

Answer 10

starting with the first screen shot, leave the second for a moment

Answer 11

yes i was about to suggest that as well
thank you

Answer 12

sounds good!

Answer 13

you have
\[f(x)=6x-x^2\] you know \(f'(x)\) ?

Answer 14

no, that is all i have.

Answer 15

if the answer is "no" that is fine, i can tell you

Answer 16

i'll take that as a no

\[f(x)=6x-x^2\\
f'(x)=6-2x\]

Answer 17

no i don't I'm sorry

Answer 18

you see it now right? do you know how i got it?

Answer 19

mmm somewhat I'm a little lost on why there is a x^2 on the answer

Answer 20

there isn't

Answer 21

the derivative of \(6x\) is \(6\) and the derivative of \(x^2\) ix \(2x\)

Answer 22

this is an online class and is extremely difficult, but yes i see the derivative now, thank you

Answer 23

you want the equation of the tangent line
you need a point and a slope
the point is \((2,8)\) from your eyeballs and the slope is \(f'(2)\)

Answer 24

okay i see, and from there you use the mx+b? to find the equation

Answer 25

no, from there you use
\[y-y_1=m(x-x_1)\] the point slope formula
then you can convert it to \(y=mx+b\)

Answer 26

you good from there? or need more help?

Answer 27

okay I'm copying down your notes, i am going to work it out to make sue its right

Answer 28

ok i can check if you like
should take a minute or two
what system ? webassign?

Answer 29

yes it is web assign

Answer 30

hows is working out?

Answer 31

i am completely lost actually, (10-8)=m(3-2)??

Answer 32

i don't know what i'm doing wrong or setting up my problem

Answer 33

whoaah nelly hold the phone
lets find \(f'(2)\) first

Answer 34

\[f'(x)=6-2x\\
f'(2)=6-4=2\]

Answer 35

point is \((2,8)\) slope is \(2\)
point slope formula gives
\[y-8=2(x-2)\] now solve for \(y\)

Answer 36

okay give me a second

Answer 37

y=2x+4

Answer 38

?

Answer 39

yup

Answer 40

that is the first line
for the second one i have to scroll up but the method will be identical

Answer 41

point is \((4,8)\) slope is \(f'(4)\) go ahead

Answer 42

okay let me solve for that one give me a second I'm a little slow for math

Answer 43

take your time

Answer 44

y-8=4(x-4) solve for y here?

Answer 45

i think your slope is wrong

Answer 46

\[f'(x)=6-2x\\
f'(4)=?\]

Answer 47

instead of the 4 its another 8?

Answer 48

your slope is wrong
first off it should be clear that the slope of the second line is negative right?

Answer 49

-2

Answer 50

yay

Answer 51

yes correct it should be negative

Answer 52

\[f'(4)=6-8=-2\]

Answer 53

wow the above f(4) confused me? how to i solve for y from there then?

Answer 54

ii think something is missing here
do you understand what a derivative is? again "no" is an ok answer,

Answer 55

no, thats what I'm supposed to know for this chapter, but it always catches me off guard

Answer 56

ok it is a function "derived" from your original function, and when evaluated at a point, it gives the slope of the tangent line at that point

Answer 57

and that derived on is what we are trying to get correct?

Answer 58

your function is \[f(x)=6x-x^2\] and the derivative is
\[f'(x)=6-2x\]

Answer 59

oh no
we have that already

Answer 60

you are not looking for a derivative you are looking for the equation of a tangent line

Answer 61

they are not the same thing at all

Answer 62

but we are getting it from derivative?

Answer 63

i agree.

Answer 64

you are getting the SLOPE of the tangent line from the derivative

Answer 65

okay yes, and that is when i got the -2

Answer 66

i apologize for the confusion, i will practice my math more before asking a lot of questions

Answer 67

then the point slope formula gives
\[y-8=-2(x-4)\] etc

Answer 68

hey no problem at all

Answer 69

y=-2

Answer 70

thank you for your patience

Answer 71

i think you should get
\[y=-2x+16\]

Answer 72

y=-2x

Answer 73

+16

Answer 74

yeah that one

Answer 75

great! i did the first math wrong with my signs, thank you!

Answer 76

y=2x+4
y=-2x+16 ?

Answer 77

yw
hope it is clear
\(f'\) is a formula for the slope
to find the equation of the tangent line use the point slope formula
yeah that is right

Answer 78

yes, its very clear now! i will practice the same problem just different numbers tomorrow

Answer 79

now that i look at the problem more carefully, we didn't need the derivative at all
you have the lines in front of you
you can compute the slope from your eyeballs

Answer 80

but its good that we did the derivatives, i learned something new!

Answer 81

if you want to do the next one as well i can help you with it, but you should repost it so i don't have to keep scrolling up

Answer 82

of course

Answer 83

calc one on line at college or high school ?

Answer 84

\[f(-5)=f(3)=0\] gets rid of the line

Answer 85

college online cal 1, I'm a biochem major and get A's in all my sciences, but math has always been a struggle

Answer 86

\[f'(-1)=0\] means the tangent line is horizontal there