Gsuppose that 4 ≤ f '(x) ≤ 5 for all values of x. what are the minimum and maximum possible values of f(6) − f(1)? 24 incorrect: your answer is incorrect. ≤ f(6) − f(1) ≤

The mean value theorem is used to link the average rate of change and the derivative of a function.

The minimum and the maximum possible values of f(6) - f(1) are 20 and 25, respectively

The given parameters are:

The mean value theorem states that:

If f(x) is continuous on [a,b] and differentiable on (a,b), then there exist a number c in (a,b) such that:

To find f(6) - f(1), we set the values of a and b to be:

So, we have:

The maximum value of f'(x) is 5.

So, we have:

Multiply both sides by 5

Rewrite as:

Similarly, the minimum value of f'(x) is 4.

So, we have:

Multiply both sides by 5

Rewrite as:

So, we have:

Hence, the minimum and the maximum possible values of f(6) - f(1) are 20 and 25, respectively

Read more about mean value theorems at:

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