91Ó°ÊÓ

To find the points where the tangent line has a certain slope, we need to calculate the derivative of the function with respect to \(x\). Apply the power rule of differentiation to the given function: $$f'(x) = \frac{d}{dx} (2x^3 - 3x^2 - 12x + 4) = 6x^2 - 6x - 12$$

Horizontal tangents occur when the slope (derivative) is equal to 0. Set the derivative equal to 0 and solve for \(x\): $$6x^2 - 6x - 12 = 0$$ To solve this quadratic equation, first divide everything by 6: $$x^2 - x - 2 = 0$$ Then factor the quadratic equation: $$(x - 2)(x + 1) = 0$$ This results in two possible values of x: \(x=2\) and \(x=-1\). Now, plug these values into the original function to find the corresponding \(y\) values: \(f(2) = 2(2)^3 - 3(2)^2 - 12(2) + 4 = -12\) \(f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 4 = 7\) So, the points where the tangent line is horizontal are \((2, -12)\) and \((-1, 7)\).

To find the points where the tangent line has a slope of 60, set the derivative equal to 60 and solve for \(x\): $$6x^2 - 6x - 12 = 60$$ Subtract 60 from both sides to obtain: $$6x^2 - 6x - 72 = 0$$ Divide everything by 6: $$x^2 - x - 12 = 0$$ This quadratic equation doesn't factor easily, so we will use the quadratic formula to solve for \(x\): $$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-12)}}{2(1)}$$ Simplify: $$x = \frac{1 \pm \sqrt{49}}{2}$$ This results in two possible values of x: \(x=3\) and \(x=-2\). Now, plug these values into the original function to find the corresponding \(y\) values: \(f(3) = 2(3)^3 - 3(3)^2 - 12(3) + 4 = 19\) \(f(-2) = 2(-2)^3 - 3(-2)^2 - 12(-2) + 4 = 12\) So, the points where the tangent line has a slope of 60 are \((3, 19)\) and \((-2, 12)\).