1.Suppose that "f" is an objects position at time "x" in seconds.

x is greater than or equal to 0 AND is less than or equal to 3

Position function:

$\displaystyle f(x) = 2x^3 - 6x^2 + 4x + 5$

Velocity:

$\displaystyle f'(x) = 6x^2 - 12x + 4$

Acceleration:

$\displaystyle f''(x) = 12x - 12$

I'm asked to find when the object is stationary. Now correct me if I'm wrong but that should be when the slopes of the position function = 0 or rather the x-intercepts of the velocity function.

However i'm having trouble finding the x-intercepts

2. Find the equation of a normal line (perpendicular to a tangent line) of

$\displaystyle y = \sqrt{x} + 1/\sqrt[3]{x}$ at x = 1

What I did was solve for y which is y=2

So I have a point now (1,2)

Then I find the derivative of the function which is

y' = (1/2)(x^(-1/2)) + (-1/3)(x^(-4/3))

I then plug in 1 which is y' = 1/6 at x =1

I use the point slope equation

y - 2 = 1/6(x-1)

y = 1/6x -1/6 + 2

y = 1/6x + 11/6

Since the equation of a normal line is the negative reciprocal:

y = -1/(1/6x + 11/6)

Is this right or not I just want to confirm

Thanks in advance