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 =
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.