Answer:
Therefore the value of c is [tex]2\sqrt {2}[/tex]
Step-by-step explanation:
Mean Value Theorem:
It state that if f(x) is defined
(i) f(x) continuous on the interval [a,b]
(ii) f(x) differentiable on (a,b).
then there exist a one number c∈ (a,b) such that
[tex]f'(c)=\frac{f(b)-f(a)}{b-a}[/tex]
Here [tex]f(x)= x-\frac {5}{x}[/tex]
(i) f(x) is continuous on [2,4]
Since it is discontinuous at 0 and 0∉[2,4].
(ii) [tex]f'(x) = 1+\frac{5}{x^2}[/tex] which is exist x≠0 and 0∉(2,4).
Therefore f(x) is differentiable on (2,4)
f(x) satisfies the Mean Value Theorem,
Then there exist a number c∈(2,4) such that
[tex]f'(c)=\frac{f(4)-f(2)}{4-2}[/tex]
[tex]\Rightarrow 1+\frac{5}{c^2}=\frac{4-\frac{5}{4}-(2-\frac{5}{2})}{4-2}[/tex]
[tex]\Rightarrow 1+\frac{5}{c^2} = \frac{2+\frac{5}{4}}{2}[/tex]
[tex]\Rightarrow 1+\frac{5}{c^2}=1+\frac{5}{8}[/tex]
[tex]\Rightarrow c^2 = 8[/tex]
[tex]\Rightarrow c= 2\sqrt {2}[/tex]
Therefore the value of c is [tex]2\sqrt {2}[/tex]
