Let f ( t ) be the weight (in grams) of a solid sitting in a beaker of water. suppose that the solid dissolves in such a way that the rate of change (in grams/minute) of the weight of the solid at any time t can be determined from the weight using the forumula f'(t)=-4f(t)(1+f(t))
if there is 5 grams of solid at time t=2 estimate the amount of solid 1 second later.

3 grams

Step-by-step explanation:

first of all

the weight in grams has been given to be f (t)

since there are 5 grams of solid in the beaker

then;

but at a time t = 2

(t represents time)

the expression for the rate of change (in grams/minute)

subtituting the value of f(t) into the expression above

we already showed that f (2) = 5, substituting this into the rate of change expression

converting this into second, we will need to divide by 60 second

the weight is decreasing 2 g/s at a time 2 ( the minus sign shows decrease)

the amount of solid 1 second later will be

the amount of solid in the beaker subtract the amount of solid that dissolved

5 - 2 = 3 grams

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option