Question

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.

Answer 1

Answer:

Therefore the value of c is $2\sqrt5$.

Step-by-step explanation:

Mean value Theorem:

Let a function f:[a,b]$\rightarrow \mathbb{R}$ be such that

1. f is continuous on [a,b], and
2. f is differentiable at every point of (a,b).

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

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

Given function is

$f(x)=x-\frac 4x$

1.

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

f is continuous on [4,5]

2.

$f'(x)=1+\frac{4}{x^2}$ 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

$f'(c)=\frac{f(5)-f(4)}{5-4}$

$\Rightarrow 1+\frac{4}{c^2}=\frac{(5-\frac45)-(4-\frac44)}{5-4}$     [ plugging x= c in f'(x) to find f'(c)]

$\Rightarrow 1+\frac{4}{c^2}=\frac{(5-\frac45-4+1)}{1}$

$\Rightarrow \frac{4}{c^2}=2-\frac45-1$

$\Rightarrow \frac{4}{c^2}=1-\frac45$

$\Rightarrow \frac{4}{c^2}=\frac15$

$\Rightarrow c^2=20$

$\Rightarrow c=\pm2\sqrt5$

Since $c=-2\sqrt5\notin$ (4,5)

$\therefore c=2\sqrt5$