Find each x-value at which the tangent line to f/left (x/right) = x^3-6x^2-34x+40 f (x) = x 3 - 6 x 2 - 34 x + 40 has slope 2.

x ∈ {-2, 6}

Step-by-step explanation:

You want the solutions to ...

f' (x) = 2

Taking the derivative, you get ...

f' (x) = 3x^2 - 12x - 34 = 2

Subtract 2 and factor:

3x^2 - 12x - 36 = 0 = 3 (x^2 - 4x - 12) = 3 (x - 6) (x + 2)

The values of x that satisfy this equation are the values of x that make the factors be zero: x = 6, x = - 2.

The slope of a tangent to the curve will be 2 at x=-2 and at x=6.