Question

for f(x)=3x^4-12x^3+6

a) find f'(x)
b) at x=1, the slope of the graph of f is ____ ?
c) at x=1, the equation of the tangent line is y=____?
d) the tangent line is horizontal at x=___?

someone please explain to me how to solve this.

Answer 1

Ok show me what you have done so far.

Answer 2

\[f(x)= 3x ^{4}-12x ^{3}+6 \] is a better view of the question

Answer 3

\[3(x ^{4}-4x ^{3}+2) =\]

Answer 4

I am not really sure where to go from there

Answer 5

for the f'(x) i got 12x^2(x-3)

Answer 6

Well here it doesn't necessarily matter if you factor out a 3 or not. I would actually just not factor the 3 out at all.
so \[f(x)=3x^4-12x^3+6\] now just differentiate each term of the polynomial.
Which is shown \[nx^\left( n-1 \right)\]

Answer 7

so 12x^3-36x^2

Answer 8

how do you figure the slope of the graph at a specific point

Answer 9

yes so
\[f'(x)=12x^3-36x^2\] or how you factored it
\[f'(x)=12x^2(x-3)\]

Answer 10

Now what is the slope of the tangent line at x=1?

Answer 11

yes it says at x=1 what is the slope of f

Answer 12

do I graph this to help me?

Answer 13

no you solve this analytically

Answer 14

find \[f'(1)=12x^3-36x^2\]

Answer 15

-24

Answer 16

then what is the equation of the tangent line at x=1

Answer 17

ah i see where that inputting the one comes in. that makes sense

Answer 18

Ok so you have to use point-slope form. You are given the slope of the line at x=1 you are given the point x=1 now all you must find is the corresponding y value by plugging in x=1 into the original function.

Answer 19

ah -3 right

Answer 20

So (1,-3) m=-24 into point slope

Answer 21

Oh i thought you mean the original function like the given question... ah

Answer 22

Wait did i confuse you on a step?

Answer 23

we would be using y=mx+b correct?

Answer 24

Well you could.

Answer 25

so b is 21

Answer 26

yep

Answer 27

awesome thank you!!! great help

Answer 28

np glad to help :)

Answer 29

Also the tangent line is horizontal when the f'(x) = 0
so
\[12x^3-36x^2=0\]