Position, Velocity & maybe Acceleration using Derivatives

Question

1. Find the points on the graph of f(x)=1/3x^3+x^2-x-1 at which slope is a)2 and b)0?

 

2. Give a position function s(t)=t+1/t+1. Find the equation of the velocity v(t).

Answer 1

1. Find the points on the graph of f(x) = (1/3)x³ + x² - x - 1 at which slope is a)2 and b)0?

 

     df(x)/dx = x² + 2x - 1

 

a)  2 = x² + 2x - 1

     0  = x² + 2x - 3

     0 = (x+3)(x-1)

     x = -3 and 1

 

b)  0 = x² + 2x - 1

     This one doesn't factor, so use the quadratic formula (look it up). 

     Answer is x = -1 ± √2

 

2)  You didn't use parentheses so your equation is ambiguous as written.  I think it's

 

     s(t) = t + 1/(t+1) = t + (t+1)^-1

 

     The velocity is:

 

     v(t) = ds(t)/dt

 

     Can you finish it from here?  Use the Power and Chain Rules.

Answer 2

1)

 

Set the derivative of f(x) to 2.

 

f'(x) = 2

 

x² + 2x - 1 = 2

 

x² + 2x - 3 = 0

 

 

Use the quadratic formula or factoring using FOIL to solve for x.  Then evaluate f(x) at that x value to find the points.  You will have 2 points.

 

 

When slope is zero, you are finding your extreme values.

 

f'(x) = 0

 

x² + 2x - 1 = 0

 

 

Again, use the quadratic formula to solve for x.  Then evaluate f(x) at the x values.  You will have 2 extreme values.

 

2)

 

Take the derivative of s(t).