Assuming [tex]f[/tex] is differentiable over the interval (1, 5), by the mean value theorem we know that there is some [tex]1<c<5[/tex] such that
[tex]f'(c)=\dfrac{f(5)-f(1)}{5-1}[/tex]
We're given that [tex]2\le f'(x)\le3[/tex] for all [tex]x[/tex], so
[tex]2\le\dfrac{f(5)-f(1)}4\le3\implies8\le f(5)-f(1)\le12[/tex]
that is, the minimum possible value is 8, and the maximum possible value is 12.
The question is an illustration of mean value theorem.
The minimum and the maximum values are: 8 and 12, respectively.
The given parameters are:
[tex]\mathbf{2 \le f'(x) \le 3}[/tex]
Mean value theorem states that:
[tex]\mathbf{f'(c) = \frac{f(b) -f(a)}{b -a}}[/tex]
In this case: a = 1, b = 5
So, we have:
[tex]\mathbf{f'(c) = \frac{f(5) -f(1)}{5 -1}}[/tex]
[tex]\mathbf{f'(c) = \frac{f(5) -f(1)}{4}}[/tex]
Multiply both sides by 4
[tex]\mathbf{4f'(c) = f(5) -f(1)}[/tex]
Rewrite as:
[tex]\mathbf{f(5) -f(1) = 4f'(c) }[/tex]
Substitute x for c
[tex]\mathbf{f(5) -f(1) = 4f'(x) }[/tex]
When f'(x) = 2, we have:
[tex]\mathbf{f(5) -f(1) = 4 \times 2 = 8 }[/tex]
When f'(x) = 3, we have:
[tex]\mathbf{f(5) -f(1) = 4 \times 3 = 12 }[/tex]
So, the minimum and the maximum values are: 8 and 12, respectively.
Read more about mean value theorems at:
https://brainly.com/question/19052862
