find the area between the following function and the line tangent to the curve at the given value.?

I'll get you started, but what you're asking seems incomplete as you need a third boundary to truly answer this question.

To find the tangent line equation:

slope = dy/dx(f(x)) = -3(3-x)^2, dy/dx(1) = -3(3-1)^2 = -12

Need 1 point: at x = 1, y = (3-1)^3 = 8

y-y1=m(x-x1), y-8 = (-12)(x-1)

y=-12x+20.

Now you have two equations, and you'll need a third boundary (probably the x axis) to solve for the area between the functions. To do that, set up a double integral that starts at your tangency function and ends at your initial function in the x direction, and from 0 to your max limit in the y direction (they intersect at 1,8, so your max y is 8).

y = (3 - x)³

dy/dx = 3(3 - x)²(- 1)

dy/dx = - 3(3 - x)²

At x = 1,

y = (3 - 1)³

y = 2³

y = 8

dy/d(1) = - 3(3 - 1)²

dy/d(1) = - 3(2)²

dy/d(1) = - 3(4)

dy/d(1) = - 12

m = dy/d(1), so

m = - 12

b = y - mx

Subbing known values,

b = 8 - [- 12(1)]

b = 8 - (- 12)

b = 8 + 12

b = 20

Equation of Tangent Line:

y = - 12x + 20

¯¯¯¯¯¯¯¯¯¯¯¯

The area cannot be determined

because the tangent intersects

the curve at only one point, so

there cannot be an interval of

integration.