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G suppose 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) ≤

Respuesta :

LammettHash
By the mean value theorem, given any interval [tex][a,b][/tex] in the domain of [tex]f(x)[/tex], which would be [tex](-\infty,\infty)[/tex] because we know the derivative exists everywhere, we know there is some [tex]c\in(a,b)[/tex] such that

[tex]f'(c)=\dfrac{f(b)-f(a)}{b-a}[/tex]

Let [tex][a,b]=[1,6][/tex].

At its highest, we're given that [tex]f'(c)=5[/tex], so it follows that the largest value of [tex]f(6)-f(1)[/tex] would be

[tex]5=\dfrac{f(6)-f(1)}{6-1}\implies f(6)-f(1)=25[/tex]

At its lowest, we would have [tex]f'(c)=4[/tex] and so

[tex]4=\dfrac{f(6)-f(1)}{6-1}\implies f(6)-f(1)=20[/tex]

Thus [tex]20\le f(6)-f(1)\le25[/tex].
MrRoyal

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

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