^{3}(2x+1)

^{4}at x= -1

state the equation of the line tangent to the graph of f(x)= (2-x)^{3}(2x+1)^{4} at x= -1

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The first step is to find the general form of the slope of f(x). We can find this by differentiating f(x):

f'(x) = -(x-2)^2*(2x+1)^3*(14x-13)

Note: At any given x, f'(x) will give us the slope of the tangent line of f(x) at that particular point.

Let y = f(x) and m = f'(x). At x=-1 then,

y = f(-1) = (2-(-1))^{3}(2(-1)+1)^{4 }= 27

m = f'(-1) = -((-1)-2)^2*(2(-1)+1)^3*(14(-1)-13) = -243

A line takes the general form:

y = mx + b

plugging in the values...

(27) = (-243)(-1) + b

b = -216

Hence, the equation of the line tangent to the graph of f(x) is

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