Mathematics, 19.09.2019 08:00 brucewaynel1043

Gsuppose 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) ≤

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Answer from: timozy95

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:

$\mathbf{4 \le f'(x) \le 5}$

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:

$\mathbf{f'(c) = \frac{f(b) - f(a)}{b - a}}$

To find f(6) - f(1), we set the values of a and b to be:

$\mathbf{a = 1}$

$\mathbf{b = 6}$

So, we have:

$\mathbf{f'(c) = \frac{f(b) - f(a)}{b - a}}$

$\mathbf{f'(c) = \frac{f(6) - f(1)}{6 - 1}}$

$\mathbf{f'(c) = \frac{f(6) - f(1)}{5}}$

The maximum value of f'(x) is 5.

So, we have:

$\mathbf{5 = \frac{f(6) - f(1)}{5}}$

Multiply both sides by 5

$\mathbf{5 \times 5= \frac{f(6) - f(1)}{5} \times 5}$

$\mathbf{25= f(6) - f(1)}$

Rewrite as:

$\mathbf{f(6) - f(1) = 25}$

Similarly, the minimum value of f'(x) is 4.

So, we have:

$\mathbf{4 = \frac{f(6) - f(1)}{5}}$

Multiply both sides by 5

$\mathbf{5 \times 4= \frac{f(6) - f(1)}{5} \times 5}$

$\mathbf{20= f(6) - f(1)}$

Rewrite as:

$\mathbf{f(6) - f(1) = 20}$

So, we have:

$\mathbf{20 \le f(6) - f(1) \le 25}$

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:

link

Answer from: bridgette143375

By the mean value theorem, given any interval [a,b]

in the domain of f(x)

, which would be $(-\infty,\infty)$ because we know the derivative exists everywhere, we know there is some $c\in(a,b)$ such that

$f'(c)=\dfrac{f(b)-f(a)}{b-a}$

Let [a,b]=[1,6]

.

At its highest, we're given that f'(c)=5

, so it follows that the largest value of f(6)-f(1)

would be

$5=\dfrac{f(6)-f(1)}{6-1}\implies f(6)-f(1)=25$

At its lowest, we would have f'(c)=4

and so

$4=\dfrac{f(6)-f(1)}{6-1}\implies f(6)-f(1)=20$

Thus $20\le f(6)-f(1)\le25$ .

Answer from: Quest

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Answer from: Quest

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