Respuesta :
Answer:
Therefore the value of c is [tex]2\sqrt5[/tex].
Step-by-step explanation:
Mean value Theorem:
Let a function f:[a,b][tex]\rightarrow \mathbb{R}[/tex] be such that
- f is continuous on [a,b], and
- f is differentiable at every point of (a,b).
Then there exists at least a point c in (a,b) such that
[tex]f'(c)=\frac{f(b)-f(a)}{b-a}[/tex]
Given function is
[tex]f(x)=x-\frac 4x[/tex]
1.
f is continuous on its domain, which includes [4,5]
f is continuous on [4,5]
2.
[tex]f'(x)=1+\frac{4}{x^2}[/tex] 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
[tex]f'(c)=\frac{f(5)-f(4)}{5-4}[/tex]
[tex]\Rightarrow 1+\frac{4}{c^2}=\frac{(5-\frac45)-(4-\frac44)}{5-4}[/tex] [ plugging x= c in f'(x) to find f'(c)]
[tex]\Rightarrow 1+\frac{4}{c^2}=\frac{(5-\frac45-4+1)}{1}[/tex]
[tex]\Rightarrow \frac{4}{c^2}=2-\frac45-1[/tex]
[tex]\Rightarrow \frac{4}{c^2}=1-\frac45[/tex]
[tex]\Rightarrow \frac{4}{c^2}=\frac15[/tex]
[tex]\Rightarrow c^2=20[/tex]
[tex]\Rightarrow c=\pm2\sqrt5[/tex]
Since [tex]c=-2\sqrt5\notin[/tex] (4,5)
[tex]\therefore c=2\sqrt5[/tex]
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 is continuous on [a,b]
- f is differentiable at every point of (a,b).
- There exists at least a point c in (a,b) such that;
f'(c) = [f(b) - f(a)]/(b - a)
- Now, we want to find all values of c in the interval [4,5] that satisfy the conclusion of the Mean-Value Theorem. Then; (a, b) = (5, 4)
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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