Question

I really need help figuring out how to do this i can't keep relying on open study lol plz help

Find the x-coordinate of each point where the graph of the function

f(x)=x^2(2x+3)^2

has a horizontal tangent line

Answer 1

You know how to find the derivative of the function?

Answer 2

ya

Answer 3

Okay, that would be the first step.
The derivative is the slope of the tangent. For what slope would we have a horizontal line?

Answer 4

y=something

Answer 5

do i use product rule and chain rule ?

Answer 6

to find the derivative, yes.

and yeah, that would be the equation itself... the slope of that horizontal line is 0.

Answer 7

so i need to let u=2x+3 and f(u)=2x^2(2x+3)^2

Answer 8

oops f(u) should be 2x^2(u)^2

Answer 9

and i then just apply product rule?

Answer 10

so here is what I've done with chain rule

u=2x+3
f(u)=2x^2(u)^2

so, dy/dx=dy/du(du/dx)

((4x)(2(u)))(2)

((4x)(4x+6))(2)

(16x^2+24x)(2)

32x^2+48x

Answer 11

now do i just plug that into product rule with the rest of the equation?

Answer 12

we'd use product rule first... since this would be the product of two functions, f(x) = x^2 and g(x) = (2x + 3)^2
Derivative of f(x)*g(x) = f'(x) g(x) + f(x) g'(x).
We need chain rule only for the derivative of g(x)

Answer 13

oh ok, that makes since i feel retorted i should have known that.

Answer 14

its fine, we learn from our mistakes. :)

Answer 15

then (x^2)d/dx(2x_3)^2+(2x+3)d/dx(x^2)

Answer 16

oops it should be (x^2)d/dx(2x_3)^2+(2x+3)^2d/dx(x^2)

Answer 17

and i only use chain rule to find out the derivative of (2x+3)^2 right

Answer 18

right.

Answer 19

k thanks I'm gonna see how all that works out for me

Answer 20

or if you really wanted, you could multiply out your f(x) and then take the derivative only using power rule, since we're going to simplify it into a polynomial form anyways.

Answer 21

that way would actually be easier, i just didnt realize it. :P

Answer 22

oh well I've actually already done it using chain rule.. but i came up with
(x^2)(8x+12)^2+(2x)(2x+3)^2

Answer 23

(8x+12)^2 i dont agree with

the g(x) = (2x + 3)^2, with the outer function being x^2 and the inner being 2x+3...

Answer 24

well i got the 8x+12 using chain rule haha

Answer 25

u=2x+3
h(u)=u^2
=(2u)(2)
=(2(2x+3))(2)
=(4x+6)2
=8x+12

Answer 26

okay, that parts correct. you just squared your answer in the equation, which didn't make sense...

Answer 27

ya because its squared in the original equations x^2(2x+3)^2

Answer 28

it is squared there, but the derivative of that part is not.
8x + 12 is the exact derivative of g(x), so back in the product rule, you'd insert it exactly as such, the derivative of g(x).

Answer 29

ok so it is (x^2)(8x+12)+(2x)(2x+3)

Answer 30

(x^2)(d/dx((2x+3)^2)) + d/dx( x^2 )(2x + 3)^2 = (x^2)(8x+12)+(2x)(2x+3)^2
^----------------equals----------------^

the (2x + 3) would be squared still, since that's the original equation I believe you were thinking of.

Answer 31

ya ok that makes sense so only the term that we took the derivative of loses its square

Answer 32

essentially, yes

Answer 33

ok so now that we have the product rule set out once i solve for that how do i get a final numerical answer

Answer 34

well, since we now have the derivative function, we'll go back to that "horizontal tangent" thing. We want to find the points where the derivative, or slope of the tangent, is 0. So, we can just set this up by setting the derivative equal to 0 and solving for x.

(x^2)(8x+12)+(2x)(2x+3)^2 = 0 We'll need to simplify the left-hand side first, getting it into standard polynomial form. So, we'd multiply our the (2x + 3)^2, etc.

Answer 35

2x(2x + 3)^2 + (x^2)(8x + 12)
= 2x(4x^2 + 12x + 9) + (8x^3 + 12x^2)
= 8x^3 + 24x^2 + 18x + 8x^3 + 12x^2
= 16x^3 + 36x^2 + 18x

that's what I get

Answer 36

Now, if we set this equal to 0...
16x^3 + 36x^2 + 18x = 0 we can factor out a monomial (2x)
2x(8x^2 + 18x + 9) = 0 then divide off the 2, we need factors of 8*9 that add up to 18.

x(8x^2 + 18x + 9) = 0 72 factors into 1+72, 2+36, 3+24, 4+18, 6+12, and 8+9.
6+12 adds up to 18, so we'll split the 18x up and factor by grouping.

x(8x^2 + 6x + 12x + 9) = 0
x(2x(4x + 3) + 3(4x + 3)) = 0
x(4x + 3)(2x + 3) = 0 Then 0-product property, a*b = 0; a=0, b=0 to solve.
x = 0

4x + 3 = 0
4x = -3
x = -3/4

2x + 3 = 0
2x = -3
x = -3/2

And we're done! x = 0, -3/4, and -3/2 are the x-coordinates of the points where we have horizontal tangents.

Answer 37

im not sure why but thats actually what i got and al three of those answers are wrong.. at least according to my online homework..

Answer 38

nevermind man i had to add the two x coordinates together thats the right answer