Respuesta :
The MVT guarantees the existence of [tex]c\in(1,3)[/tex] such that
[tex]f(c)=\dfrac{f(3)-f(1)}{3-1}=\dfrac{f(3)-4}2[/tex]
Since [tex]f'(x)\le2[/tex] for all [tex]x\in[1,3][/tex], we have
[tex]\dfrac{f(3)-4}2\le2\implies f(3)-4\le4\implies f(3)\le8[/tex]
so that [tex]V=8[/tex].
The mean value theorem is used to link the average rate of change and the derivative of a function.
The value of V is 8.
The given parameters are:
[tex]\mathbf{f'(x) \le 2}[/tex]
[tex]\mathbf{f(1) = 4}[/tex]
[tex]\mathbf{f(3) \le V}[/tex]
[tex]\mathbf{x \in [1,3]}[/tex]
Mean value theorem states that:
If [tex]\mathbf{f(x)\ is\ continuous\ at }[/tex] [a,b] and
[tex]\mathbf{f(x)\ is\ differentiable\ on }[/tex] (a,b),
Then there is a point c in (a,b), such that:
[tex]\mathbf{f'(c) = \frac{f(b) - f(a)}{b - a}}[/tex]
From the question, we understand that: f is differentiable
This means that:
[tex]\mathbf{f'(c) = \frac{f(b) - f(a)}{b - a}}[/tex]
So, we have:
[tex]\mathbf{f'(c) = \frac{f(3) - f(1)}{3 - 1}}[/tex]
[tex]\mathbf{f'(c) = \frac{f(3) - f(1)}{2}}[/tex]
Substitute 4 for f(1)
[tex]\mathbf{f'(c) = \frac{f(3) -4}{2}}[/tex]
Recall that: [tex]\mathbf{f'(x) \le 2}[/tex]
The equation becomes
[tex]\mathbf{\frac{f(3) -4}{2} \le 2}[/tex]
Cross multiply
[tex]\mathbf{f(3) -4 \le 4}[/tex]
Add 4 to both sides
[tex]\mathbf{f(3) \le 8}[/tex]
From the question, we have: [tex]\mathbf{f(3) \le V}[/tex]
By comparisons;
[tex]\mathbf{V = 8}[/tex]
Hence, the value of V is 8.
Read more about mean value theorems at:
brainly.com/question/3957181