If f(2) = 12 and f '(x) ≥ 3 for 2 ≤ x ≤ 5, how small can f(5) possibly be?

21

Step-by-step explanation:

From our previous knowledge, if f'(x) = m ,

Then f(x) = mx+c

For a smaller value of f(5), m should be the smallest possible integer if m = 3

i.e. f(x) = 3x+c

replacing the value of x with 2 given from the question to get the value of c, we have:

f(x) = 3x+c

f(2) = (3×2)+c

where; f(2) = 12

12 = 6+c

- c = -12 + 6

- c = - 6

c = 6

Now, f(x) can now be;

f(x) = 3x + 6

f(5) = (3×5)+6

f(5) = 15 + 6

f(5) = 21

Therefore, smallest possible value of f(5) is 21

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.