If g(x) = x*f(x) where f(3) = 4 and f'(3) = -2...

Question

If g(x) = x*f(x) where f(3) = 4 and f'(3) = -2, find an equation of the tangent line to the graph of g at the point where x = 3

I get that I need to apply the product rule. I believe the derivative is 4 - 2x. However, when I pug in 3, and solve for b, I don't get the right answer

 

#47, p190

ANSWER: -2x + 18

 

Answer 1

g(x) = xf(x)       f(3) = 4       f'(3) = -2         g(3) = 3f(3) = 3(4) = 12

 

Find slope by differentiating g(x)

g'(x) = f(x) + xf'(x)           g'(3) = f(3) + 3f'(3) = 4 + 3(-2) = -2

 

Use point slope formula to get the equation of the tangent line at x =3.

y - y1 = m(x-x1)

 

g - g(3) = g'(3) (x-3)             g - 12 = -2 (x-3)         g = -2x + 18

 

Answer 2

First, we need to find f(x).  Then multiply it by x to get g(x).

 

 

f(3) = 4  indicates the coordinate point (3, 4).  The derivative of f(x) when x=3 is  -2.  The slope of the tangent line is the derivative.  Also, note that x=3 is a consistent x value.

 

So the equation of the tangent line so far knowing all this is

 

f(x) = -2x + b

 

 

Next, plug in the coordinate point (3, 4) into this equation to find b.

 

4 = -2(3) + b

 

4 = -6 + b

 

10 = b

 

f(x) = -2x + 10

 

 

g(x) = x * f(x)

g(x) = 3(-2x + 10)

g(x) = -6x + 30

 

 

Now we can find the tangent line to g(x) at x=3.  First, evaluate g(3).

 

g(3) = -6(3) + 10

g(x) = -18 + 10

g(3) = -8

 

 

The coordinate point is (3, -8).  The derivative of g(x) is -6. 

 

Tangent line to g(x):

 

y = -6x + b

 

 

Plug in the coordinate (3, -8) into the equation to find b.

 

-8 = -6(3) + b

 

-8 = -18 + b

 

10 = b

 

 

The tangent line to g(x) at x=3 is

 

y = -6 x+ 10