If f(4) = 12 and f '(x) ≥ 3 for 4 ≤ x ≤ 9, how small can f(9) possibly be?

We want to find the minimum possible value of f(9) given that:

We will see that the smallest value that f(9) can take is 27.

We know that the derivate of f(x) gives the slope to the tangent line to the graph of f(x) in a given point where it is evaluated.

If this slope is positive, the function is increasing. So if we want to find the minimum value for f(9), then we need to take the smallest slope possible (knowing that it is positive).

Then we must use f'(x) = 3

Now, we can integrate this to get:

f(x) = 3*x + b

Where b is a constant of integration, to find the value of b, we can use the fact that f(4) = 12, then:

12 = f(4) = 3*4 + b

12 = 12 + b

12 - 12 = 0  = b

Thus the equation is just:

f(x) = 3*x

Now we can evaluate this in x = 9 to get:

f(9) = 3*9 = 27

The smallest value that f(9) can take is 27.

If you want to learn more, you can read:

link

  27

Step-by-step explanation:

If the slope of f(x) is no less than 3, then the value of f(9) can be no less than 27.

  f(x) ≈ f(4) +f'(x)·(x-4)

  f(9) = 12 +3(9 -4) = 27 . . . . . for f'(x) = 3.

For larger numbers of f'(x), f(9) will be larger.

answer: -0.5 would be the most likley outcome

step-by-step explanation:

(−$11) ⋅ 12

deductions are negative and there are 12 months in a year