For the function f(x)=x−4x, find all values of c in the interval [4,5] that satisfy the conclusion of the Mean-Value Theorem. If appropriate, leave your answer in radical form. Enter all fractions in lowest terms.

Therefore the value of c is .

Step-by-step explanation:

Mean value Theorem:

Let a function f:[a,b] be such that

Then there exists at least a point c in (a,b) such that

Given function is

1.

f is continuous on its domain, which includes [4,5]

f is continuous on [4,5]

2.

which exists for all x≠0. So, x exits in (4,5).

f is differentiable at every point of (4,5).

All hypotheses of Mean Value Theorem are satisfied by this function .

So, there exits a point c such that

    [ plugging x= c in f'(x) to find f'(c)]

Since (4,5)

The value of c in the interval [4,5] that satisfy the conclusion of the Mean-Value Theorem is; c = 2√5

We are given the function;

f(x) = x - (4/x)

Now according to mean value theorem, we have the following conditions;

        f'(c) = [f(b) - f(a)]/(b - a)

Thus;

f(5) = 5 - (4/5)

f(5) = 21/5

f(4) = 4 - (4/4)

f(4) = 3

Now, when x = c, we have;

f(c) = c - (4/c)

f'(c) = 1 + (4/c²)

Thus, from  f'(c) = [f(b) - f(a)]/(b - a), we now have;

1 + (4/c²) = [(21/5) - 3]/(5 - 4)

1 + (4/c²) = 18/15

4/c² = (18/15) - 1

4/c² = 3/15

c² = (15 * 4)/3

c = √20

c = 2√5

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the answer is y < -3 or   y > 3

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