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:
[tex]\mathbf{4 \le f'(x) \le 5}[/tex]
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:
[tex]\mathbf{f'(c) = \frac{f(b) - f(a)}{b - a}}[/tex]
To find f(6) - f(1), we set the values of a and b to be:
[tex]\mathbf{a = 1}[/tex]
[tex]\mathbf{b = 6}[/tex]
So, we have:
[tex]\mathbf{f'(c) = \frac{f(b) - f(a)}{b - a}}[/tex]
[tex]\mathbf{f'(c) = \frac{f(6) - f(1)}{6 - 1}}[/tex]
[tex]\mathbf{f'(c) = \frac{f(6) - f(1)}{5}}[/tex]
The maximum value of f'(x) is 5.
So, we have:
[tex]\mathbf{5 = \frac{f(6) - f(1)}{5}}[/tex]
Multiply both sides by 5
[tex]\mathbf{5 \times 5= \frac{f(6) - f(1)}{5} \times 5}[/tex]
[tex]\mathbf{25= f(6) - f(1)}[/tex]
Rewrite as:
[tex]\mathbf{f(6) - f(1) = 25}[/tex]
Similarly, the minimum value of f'(x) is 4.
So, we have:
[tex]\mathbf{4 = \frac{f(6) - f(1)}{5}}[/tex]
Multiply both sides by 5
[tex]\mathbf{5 \times 4= \frac{f(6) - f(1)}{5} \times 5}[/tex]
[tex]\mathbf{20= f(6) - f(1)}[/tex]
Rewrite as:
[tex]\mathbf{f(6) - f(1) = 20}[/tex]
So, we have:
[tex]\mathbf{20 \le f(6) - f(1) \le 25}[/tex]
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:
https://brainly.com/question/12369096
