Suppose an object moves along the y-axis (marked in feet) so that its position at time x (in seconds) is given by the f(x) = 96x -16x^2, find the following. A) The instantaneous velocity function v = f (x).
B) The velocity when x = 0 and x = 4 seconds.
C) The time(s) when v = 0

A) v = f(x) = 96 -32x

B)the velocity when x = 0

V = 96 ft/s

The velocity when x =4

V = -32 ft/s

C). Time when v= 0

3 seconds = x

Step-by-step explanation:

f(x) = 96x -16x^2

The above equation represents the position of the object at x time along the x axis.

The velocity will be determined by differentiating the equation with respect with x

f(x) = 96x -16x^2,

D f(x)/Dx= 96 -2(16x)

D f(x)/Dx= 96 -32x

v = f(x) = 96 -32x

The velocity when x = 0

v = f(x) = 96 -32x

V = f(0) = 96-32(0)

V = 96 ft/s

The velocity when x = 4

v = f(x) = 96 -32x

V = f(4) = 96-32(4)

V = 96 - 128

V = -32 ft/s

Time when v= 0

v = f(x) = 96 -32x

0= 96 -32x

-96= -32x

-96/-32=x

3 seconds = x

the solution is 12x-6y

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so: