What is the Equation for the Tangent Line of g(x) at Point x=2?

[Johny 5]

Homework Statement

Let f be a function differentiable function with f(2) = 3 and f'(2) = -5, and let g be the function defined by g(x) = xf(x). Which of the following is an equation of the line tangent to the graph of g at point where x=2?
a. y=3x
b y-3 = -5(x-2)
c y-6 = -5(x-2)
d y-3 = -7(x-2)
e y-6 = -10(x-2)

Homework Equations

The Attempt at a Solution

given:
x=2
y=3
so
3=-5(2)+b
b=13
so
g(x) = x(-5x+13)
g(x) = -5x^2 + 13x
g'(x) = -10x +13
g'(2) = -7
point slope form:
d. y-3 = -7(x-2)
is that correct?

 

[Rake-MC]

What you've done here is taken the tangent line to f(x) at point x = 2 and assumed it to be f(x).

 

[Johny 5]

so the whole thing is incorrect?

 

[sutupidmath]

[tex]g'(x)=f(x)+xf'(x)[/tex]

[tex]g'(2)=f(2)+2f'(2)=3-10=-7[/tex] this will be the slope of the tangent line at x=2, on the graph of g. now slope-intercept formula for a line is

[tex]g-g(x_0)=k(x-x_0)=>g-g(2)=-7(x-2)[/tex] and youre almost done.

Just find what g(2) is, substitute it, and bummmm, that's your answer.

 

[Rake-MC]

Well you tell me. Firstly you did not finish solving your equation, you stopped as soon as you saw that g'(2) = -7 and assumed that was the answer. If you had solved it all you would see that you get y - 6 = -7(x - 2). This is not an option.

I will however give you a starter:

y = -5x + 13 is NOT the function f(x), it is the equation of a tangent and only applies as far as we know at point x = 2.
Edit: Scratch that - stupid error.
Read what sutupidmath said.

 

[Johny 5]

g(2) = (2)f(2)
g(2) = 2*3
g(2) = 6... hmm that's not an option

 

[Johny 5]

I stopped at g'(2) = -7 because that would be the slope, m, (you got it too) and i just plugged the given into the point slope formula (x = 2 and y = 3)

 

[sutupidmath]

g(2) = (2)f(2)
g(2) = 2*3
g(2) = 6... hmm that's not an option

Hmmm... i just glanced at the options now... i don't know, as far as i am concerned, if the problem is exactly phrased like this, then i guess they should put this one as an option too. Are u sure that you have written everything correctly?

 

[sutupidmath]

I stopped at g'(2) = -7 because that would be the slope, m, (you got it too) and i just plugged the given into the point slope formula (x = 2 and y = 3)

Your y coordinate won't be 3, since you are trying to find the slope of the tangent line at x=2 on the graph of g, so your point actually will be (2,g(2)) which is (2, 6). At least this is what you wrote on the problem on the first post.

 

[Johny 5]

Hmmm... i just glanced at the options now... i don't know, as far as i am concerned, if the problem is exactly phrased like this, then i guess they should put this one as an option too. Are u sure that you have written everything correctly?

yup the problem is word for word from my worksheet... (i just re-read)

 

[Rake-MC]

[...] point slope formula (x = 2 and y = 3)

However, this is the formula for f(x) not g(x) no?

 

[sutupidmath]

yup the problem is word for word from my worksheet... (i just re-read)

HOnestly, it might also be since it is 4 in the morning here, that i am missing sth really obvious there, or i really think that there is some kind of missinterpretation of that problem. Since there is no other way of goin about it, as far as i am concerned. so just add another
option

f)y-6=-7(x-2)

 

[Rake-MC]

HOnestly, it might also be since it is 4 in the morning here, that i am missing sth really obvious there, or i really think that there is some kind of missinterpretation of that problem. Since there is no other way of goin about it, as far as i am concerned. so just add another
option

f)y-6=-7(x-2)

But that would make three of us reading it incorrectly, and it's not 4am here..

 

[Johny 5]

it's for xf(x) (which is g(x)) or at least i thought it was... but you pointed out what i wrote as a function of f(x) is not f(x) its just for f(x) at x=2

 

[sutupidmath]

But that would make three of us reading it incorrectly, and it's not 4am here..

Well, with this i just wanted to leave a 0.01% chance that i am getting the whole thing wrong. because i am quite sure that the way i did it is the correct one, because it seems a quite straightforward problem.

 

[sutupidmath]

it's for xf(x) (which is g(x)) or at least i thought it was... but you pointed out what i wrote as a function of f(x) is not f(x) its just for f(x) at x=2

So, are u sayin' now that you want to find the tangent line of f(x) at x=2, or what?

if so then it would be b) y-3 = -5(x-2)

 

[Rake-MC]

So, are u sayin' now that you want to find the tangent line of f(x) at x=2, or what?

if so then it would be b) y-3 = -5(x-2)

That's what I got too, but I do believe that they want the tangent of g(x) at x=2. For which we can't get a answer that complies with the options

 

[sutupidmath]

That's what I got too, but I do believe that they want the tangent of g(x) at x=2. For which we can't get a answer that complies with the options

We cannot get an answer that complies withthe options because there is no such answer!

 

[Johny 5]

so the answer is y-6 = -7(x-2)?... which isn't an option...

 

[Rake-MC]

as far as me and sutupidmath can see, yes.

 

[Johny 5]

i just graphed the function that we got to be g(x), which is xf(x)
C.y-6 = -5(x-2) was a tangent to the line y=x(-5x+13) at x=2 (2,6)

NOTE: y-6 = -7(x-2) also got the same results at x=2??!? this is the equation that wasn't an option... the other one is but i don't know how to get there

 

[sutupidmath]

i just graphed the function that we got to be g(x), which is xf(x)

First i don't know how could you graph g(x) when we don't actually know at first place what f(x) is, we could try to figure that out, but still.

C.y-6 = -5(x-2) was a tangent to the line y=x(-5x+13) at x=2 (2,6)

I am completely confused: A tangent to the line :y=x(-5x+13) , where did u get this one from? Or, just don't tell me that your f(x) =-5x+13, and u didn't tell us this from the very begginning?

NOTE: y-6 = -7(x-2) also got the same results at x=2??!? this is the equation that wasn't an option... the other one is but i don't know how to get there

 

[sutupidmath]

if that's the case then, [tex]g(x)=-5x^2+13x[/tex] so [tex]g'(x)=-10x+13[/tex] so [tex]g'(2)=-7[/tex]

[tex]g(2)=6[/tex] which would give us again exactly the same thing. since

[tex]g-g(2)=-7(x-2)[/tex]

 

[Rake-MC]

What he has written is incorrect. y=-5x+13 is the equation for the tangent of f(x) ONLY at point x=2 (as far as we know).

[tex] f(2) = 3 [/tex]

[tex] f'(2) = -5 [/tex]

[tex] y = mx + c [/tex]

[tex] 3 = -5(2) + c [/tex]

[tex] c = 13 [/tex]

tangent for f(x) at point x=2 is: [tex] y = -5x +13 [/tex]

I believe he has become confused and thought this is the equation for the function f(x) and worked from there.