Question

verify cauchy's mean value theorem for the functions f(x) =1/x^2 and g(x) =1/x in [a, b] , a>0. show that the point c is harmonic mean of a and b.​

Answer 1

Answer:

C=a+b/2(ab) €(a,b).

Step-by-step explanation:

Given that

f(x)=1/x^2. & g(x)=1/x. in [a,b].

Thus f(x) & g(x) is continuous in [a,b].

f(x)=1/x^2

f'(x)=-2x^-3

g(x)=-x^-2

:. f(x) &g(x) are differentiable on [a,b].

Here g'(x) is nit equal to 0.

Three conditions of cauchy's mean value theorem are verified.

Then their exits a point C€(a,b) such that :

f(b)-f(a)÷g(b)-g(a)=f'(c)/g'(c)

1/b^2-1/a^2÷1/b-1/a=-2c^-3/-c^-2

a^2-b^2/(ab)^2÷a-b/ab=2c^-

a^2-b^2/ab×1/a-b=2c^-

2c^-=a+b/ab

C=a+b/2(ab) €[ a,b].

Answer 2

Answer:

the eugen values of the unit matrix order od 3