Need help finding equation of Tangent Line

[dubb]

Hello,
Here is the question from text:
Suppose the line tangent to the graph of f at x = 2 is y = 4x + 1 and suppose y = 3 x - 2 is the line tanget to the graph of g at x = 2. Find the line tangent to the following curves at x = 2.

a) y = f(x)g(x)

I posted a photo of my work. Thanks so much for your help.

 

[HallsofIvy]

The tangent line to a curve, y= f(x), at a given x= a must have the same value at x= a as f, f(a), and its slope must be equal to f'(a). If y= 4x+ 1 is tangent to y= f(x) at x= 2, then f(2)= 4(2)+ 1= 9 and f'(2)= 4. If y= 3x- 2 is tangent to y= g(x) at x= 2 the g(2)= 3(2)- 2= 4 and g'(2)= 3.

So what is f(x)g(x) at x= 2? What is (f(x)g(x))'= f'(x)g(x)+ f(x)g'(x) at x= 2?

I have a little difficulty reading your picture but that seems to be what you have done.

 

[dubb]

Hmmm...I'm really trying to understand and visualize the graph and everything you wrote. Thanks very much. Here is what I got.

So what is f(x)g(x) at x= 2? (9)(4)= 36

What is (f(x)g(x))'= f'(x)g(x)+ f(x)g'(x) at x= 2? 4•4 + 9•3 = 43

 

[HallsofIvy]

Hmmm...I'm really trying to understand and visualize the graph and everything you wrote. Thanks very much. Here is what I got.

So what is f(x)g(x) at x= 2? (9)(4)= 36

What is (f(x)g(x))'= f'(x)g(x)+ f(x)g'(x) at x= 2? 4•4 + 9•3 = 43

so the tangent line to y= f(x)g(x), at x= 2, is y= 43(x- 2)+ 36.

 

[dubb]

Wow, thanks so much. Is it possible to see if this is correct?

If I had these numbers instead:
tangent line of the graph of f is y=3x+2 at x=3 { f'(3)= 3 , f(3)=11}

tangent line of the graph of g is y=5x+1 at x=3 { g'(3)= 5 , g(3)=16}

f(x)g(x)= 11•16 = 176 makes point (3,176)

f(x)g(x)'= f'(3)g(16)+ f(11)g'(5) at x= 3 = 103 slope

the equation is y=103x-133 ?



for f(x)/g(x)
y = 11/16

(f(x)/g(x)') = -7/256

the equation of tangent is y = -(7/256)x + 197/256 ?

 

[HallsofIvy]

Wow, thanks so much. Is it possible to see if this is correct?

If I had these numbers instead:
tangent line of the graph of f is y=3x+2 at x=3 { f'(3)= 3 , f(3)=11}

tangent line of the graph of g is y=5x+1 at x=3 { g'(3)= 5 , g(3)=16}

f(x)g(x)= 11•16 = 176 makes point (3,176)

f(x)g(x)'= f'(3)g(16)+ f(11)g'(5) at x= 3 = 103 slope

This is badly written: f'(3)g(3)+ f(3)g'(3)= 3(16)+ 11(5)= 48+ 55= 103.

the equation is y=103x-133 ?

That looks good to me. As a check, if x= 3, y= 309- 133= 176= 11(16) and y'= 103= 3(16)+ 11(5)

for f(x)/g(x)
y = 11/16

(f(x)/g(x)') = -7/256

the equation of tangent is y = -(7/256)x + 197/256 ?

Yes, that is also correct.