let f be a differentiable function with f(2)=3 and f'(2)=-5, and let g be a function defined by g(x)=xf(x).?

Question

let f be a differentiable function with f(2)=3 and f'(2)=-5, and let g be a function defined by g(x)=xf(x). Create an equation for the line tangent to the graph of g at the point where x=2?

Answer 1

evaluate the function at x = 2

g(x) = xf(x)

g(2) = 2*f(2) = 2*3 = 6



find the derivative of g(x)

g(x) = xf(x)

g'(x) = f(x) + xf'(x)



evaluate the derivative at x = 2

g'(2) = f(2) + 2*f'(2)

= 3 + 2*-5

= 3 - 10

g'(2) = -7

this is the slope, m, of the tangent line



remember, the equation of a line is: y = mx + b

we now have x, y, and m

x = 2

y = 6

m = -7

now solve for b



y = mx + b

6 = -7(2) + b

6 = -14 + b

b = 6 + 14

b = 20



equation:

y = mx + b

y = -7x + 20

Answer 2

Ok, first of all, what do you need when you want to make an equation for a line?

y - y1 = m(x - x1)

The x value

The y value

and m (the slope)



It tells you the x value in the problem is 2, so now you just need a y value and the slope.



You're looking for the equation tangent to the graph of g.



Finding the derivative of g(x) will give you m(the slope).



In your problem, g(x) = xf(x)



Use the product rule



g'(x) = xf'(x) + f(x)



It tells you in the problem that x = 2, and that f(2) = 3 and f'(2) = -5, so plug these in



g'(x) = 2(-5) + 3



g'(x) = -10 +3



so m= -7



You now have the slope.

Now you just need the y-value for g(x)



g(x) = xf(x)

It tells you in the problem that x=2 and f(2) = 3, so plug these in

g(x) = 2(3)

so g(x) = 6 is your y-value, and 2 is your x, value



Your equation will be y - 6 = -7(x - 2)

Simplify and you have y = -7x + 20

Answer 3

Let the equation of the tangent line at x = 2 be:



y - y1 = m*(x - x1) ......... (1)



We use the point-slope form of a straight line here.



Now, slope of the tangent to g(x) at x = 2 equals the value of the derivative of g(x) at x = 2.

Thus, m = g'(2) .................(2)



Given:



g(x) = x*f(x)



Using the product rule of derivatives (i.e. (uv)' = u'v + uv') we have:



g'(x) = x*f'(x) + 1*f(x)

=> g'(2) = 2*f'(2) + f(2)

=> g'(2) = 2*(-5) + 3

=> g'(2) = -7 .............................(3)



Plugging this value in eqn. (1), we have:



y - y1 = -7*(x - x1) ..............(4)



Next, we find the values of x1 and y1 at x = 2, i.e. the point (x1, y1).

We know that x1 = 2, and to find y1, we have:



y1 = g(2) = 2*f(2) = 2*3 = 6.



Putting these values in eqn. (4), we have:



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

=> y - 6 = 14 - 7x

=> 7x + y - 20 = 0



which is the equation of the tangent to g(x) at x = 2.