# Suppose that f(x)= 2x^2(5-2x)3 (A) Find an equation for the tangent line to the graph of f at x=2. Tangent line: y = (B) Find the average of all values of x where the tangent line is horizontal....

Suppose that f(x)= 2x^2(5-2x)3

(A) Find an equation for the tangent line to the graph of f at x=2.

Tangent line: y =

(B) Find the average of all values of x where the tangent line is horizontal. If there are no such values, enter -1000.

Average of x values =

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You need to remember what equation of tangent line to a curve at a point is:

y - y_0 = f'(x_0)(x - x_0)

The problem provides x_0 coordinate of the point of tangency, hence, you need to find y_0 substituting 2for x in equation y=2x^2(5-2x)^3 such that:

f(2)=2*2^2(5-4)^3 => f(2) = 8

You need to differentiate the function with respect to x such that:

f'(x) = (2x^2(5-2x)^3)'

You need to use product rule to differentiate the function such that:

f'(x) = (2x^2)'(5-2x)^3 + 2x^2*((5-2x)^3)'

f'(x) = 4x(5-2x)^3 - 12x^2(5-2x)^2

Factoring out 4x(5-2x)^2 yields:

f'(x) = 4x(5-2x)^2*(5 -2x - 3x)

f'(x) = 4x(5-2x)^2*(5 -5x)

f'(x) = 20x(5-2x)^2*(1-x)

You need to substitute 2 for x in f'(x) such that:

f'(2) = 40(5-4)^2*(1-2)

f'(2) = -40

You need to substitute 2 for x_0 , 8 for y_0 and -40 for f'(x_0) in equation of tangent line such that:

y - 8 = -40(x - 2)

You need to use the slope intercept form of equation of tangent line, hence, you need to isolate y to the left side such that:

y = -40x + 80 + 8

y = -40x + 88

**Hence, evaluating the equation of the tangent line to the graph of given function, at x=2, yields y = -40x + 88.****The tangent line is horizontal if the slope m is equal to zero, which is not the case.**