Suppose f is a differentiable function such that f′(x)≤2 for all x∈[1,3]. If f(1)=4, the Mean Value Theorem says that f(3)≤V for what value of V? (Choose V as small as possible.)

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

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