Question

an object moves along the y - axis (marked in feet) so that its position at time x (in seconds) is given by the function f(x)=x^3 - 12x^2 + 36x.
a. find the instantaneous velocity function v = f(x).
b. find the velocity when x = 1.
c. find the time(s) when v = 0.

Explanation:

Step1: Differentiate to find velocity function

The derivative of $f(x)=x^{3}-12x^{2}+36x$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ is $v = f'(x)=3x^{2}-24x + 36$.

Step2: Evaluate velocity at $x = 1$

Substitute $x = 1$ into $v=f'(x)$. So $v=f'(1)=3(1)^{2}-24(1)+36=3 - 24 + 36=15$ feet per second.

Step3: Solve for $x$ when $v = 0$

Set $v = 3x^{2}-24x + 36=0$. Divide through by 3 to get $x^{2}-8x + 12=0$. Factor the quadratic equation: $(x - 2)(x - 6)=0$. Then $x=2$ or $x = 6$ seconds.

Answer:

a. $v = 3x^{2}-24x + 36$
b. 15 feet per second
c. $x = 2$ seconds and $x = 6$ seconds