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 - 15x^2 + 72x. 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: Apply power - rule for differentiation

The power - rule states that if $y = x^n$, then $y^\prime=nx^{n - 1}$. Given $f(x)=x^{3}-15x^{2}+72x$, differentiating term by term:
For the first term $x^{3}$, its derivative is $3x^{2}$; for the second term $-15x^{2}$, its derivative is $-15\times2x=-30x$; for the third term $72x$, its derivative is 72. So $v = f^\prime(x)=3x^{2}-30x + 72$.

Step2: Evaluate the velocity at $x = 1$

Substitute $x = 1$ into $v=f^\prime(x)$.
$v=f^\prime(1)=3\times1^{2}-30\times1 + 72=3-30 + 72=45$ feet per second.

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

Set $v = 3x^{2}-30x + 72=0$. First, divide the entire equation by 3: $x^{2}-10x + 24=0$.
Factor the quadratic equation: $(x - 4)(x - 6)=0$.
Set each factor equal to zero: $x-4=0$ gives $x = 4$; $x - 6=0$ gives $x = 6$.

Answer:

a. $v = 3x^{2}-30x + 72$
b. 45 feet per second
c. $x = 4$ seconds and $x = 6$ seconds