Question

Let h(x)=5−3x^3,
h'(3)=

Answer 1

@hartnn

Answer 2

could you first find h'(x) by differentiating h(x) ?

Answer 3

Use this to find the equation of the tangent line to the curve y=5−3x3 at the point (3−76) and write your answer in the form:
b) y=mx+b, where m is the slope and b is the y-intercept.

Answer 4

yes inverse function isnt

Answer 5

i got y=(-x+5)?

Answer 6

inverse ? no...
can you differentiate 5- 3x^3 ?

Answer 7

mmmm.. no. i have no idea. :/

Answer 8

aren't you just looking for the derivative of h?

Answer 9

if you haven't learned derivatives, how are you attempting this question ?

Answer 10

anyways, you'll need this : \(\dfrac{d}{dx}x^n=nx^{n-1}\)

Answer 11

im confused already. I want to learn.

Answer 12

how do i apply this equation to the question

Answer 13

\(\dfrac{d}{dx}x^n=nx^{n-1} \\ \dfrac{d}{dx}x^3=...?\)

Answer 14

nx^3-1?

Answer 15

note that n=3

Answer 16

yes.

Answer 17

so, \(\dfrac{d}{dx}x^3=3x^{2}\)
gott his ?

Answer 18

,mmmmm so answer is 3x^2? how come?

Answer 19

i'll use a shorthand notation
(x^3)' = 3x^2

Answer 20

thats not final answer...just (x^3)'

Answer 21

ok! i got it. so (x^3)'=3x^2

Answer 22

also derivative of constant =0
so, (5)'=0

Answer 23

mmm ok.

Answer 24

so, \((5-3x^3)'=0-3(3x^2)=...?\)

Answer 25

im taking note hold on

Answer 26

why its zero, is it beacuse constant?

Answer 27

yes, any constant (say c) can be written as , \(cx^0\)
and if you use the formula for x^n,
\((cx^0)'=c (0x^{0-1})=0\)
got this ?

Answer 28

YES! i think so. im taking note hold on please

Answer 29

ok.

Answer 30

Ok, so the 9x^2?

Answer 31

-9x^2

Answer 32

i forgot to write -.

Answer 33

and whats the next step?

Answer 34

so, h'(x) = -9x^2
to get h'(3) just put x=3 in that!

Answer 35

ohh cool so -81?

Answer 36

i got your message, i will wait here

Answer 37

yes, -81 is correct.
that will be the slope of your line, m=h'(3)=-81
now you just need b = y-intercept=....

Answer 38

to get that, when x= 3, what is h(3) =... ?

Answer 39

-3^3?

Answer 40

where did 5 go ?

Answer 41

note that this time its h(3) and not h'(3)

Answer 42

so, 5-3(3^3)

Answer 43

Oh why its not h'(3)?

Answer 44

i got it.

Answer 45

y= h(x) = 5-3x^3
when x=3, we need to find y
thats why.

Answer 46

that rhymed :P

Answer 47

I see

Answer 48

:)

Answer 49

so, h(3)=.. ?

Answer 50

h(3)=5-3(3^3)?

Answer 51

76?

Answer 52

5-3*27
must be negative..

Answer 53

ops-76?

Answer 54

oh.. is you '-' key of your keyboard not working? :P
just kidding..
so, you have m= -81 , x=3,y=-76
just plug in y=mx+b and find b !

Answer 55

so -167? since its -76=-243*3+b?

Answer 56

y=mx+b
-76=-81(3) +b
b= 81*3-76 = 167
O.o now you have an extra -

Answer 57

so no negative this time?

Answer 58

yeah...just 167
so your equation will be
y= -81x+167

Answer 59

I entred it but it told me wrong :(

Answer 60

let me go through all steps again...

Answer 61

Use this to find the equation of the tangent line to the curve y=5−3x^3 at the point (3−76) and write your answer in the form:

Answer 62

couldn't find any error :\
y= -81x+167
must have worked...

Answer 63

I am so sorry i entred 160!

Answer 64

it worked! thank yu!!

Answer 65

ok. no problem.
and welcome ^_^

Answer 66

A. Find the average velocity for the time period beginning when t=3 and lasting