an object moves along the y - axis (marked in feet) so that its position at time x (in seconds) is given by…

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 - 9x^2 + 24x. a. find the instantaneous velocity function v = f(x). b. find the velocity when x = 1. c. find the time(s) when v = 0.

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 - 9x^2 + 24x. a. find the instantaneous velocity function v = f(x). b. find the velocity when x = 1. c. find the time(s) when v = 0.

Answer

Explanation:

Step1: Differentiate the position function

The power - rule for differentiation is $\frac{d}{dx}(x^n)=nx^{n - 1}$. Given $f(x)=x^{3}-9x^{2}+24x$, then $v = f^{\prime}(x)=\frac{d}{dx}(x^{3}-9x^{2}+24x)=3x^{2}-18x + 24$.

Step2: Evaluate the velocity function at $x = 1$

Substitute $x = 1$ into $v(x)=3x^{2}-18x + 24$. So $v(1)=3(1)^{2}-18(1)+24=3-18 + 24=9$.

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

Set $v(x)=3x^{2}-18x + 24 = 0$. First, divide through by 3 to get $x^{2}-6x + 8 = 0$. Factor the quadratic equation: $(x - 2)(x - 4)=0$. Then $x-2=0$ or $x - 4=0$, so $x = 2$ or $x = 4$.

Answer:

a. $v = 3x^{2}-18x + 24$ b. $9$ c. $x = 2$ or $x = 4$