The function f (x) = x + lgx is known It is proved that f (x) = 3 has real number solution on interval (1,10) If X. Is a real solution of the equation f (x) = 3. € (k, K + 1) finding integer k value

The function f (x) = x + lgx is known It is proved that f (x) = 3 has real number solution on interval (1,10) If X. Is a real solution of the equation f (x) = 3. € (k, K + 1) finding integer k value

Because f (x) is a continuous function on (1,10)
Let g (x) = f (x) - 3, so g (x) is also a continuous function on (1,10)
G (1) = - 2, G (10) = 8, G (1) 0, so there exists m in (1,10) such that G (m) = 0
G (m) = 0, that is, f (3) - 3 = 0
Proof of the original formula

If f (x) = f (1 / x) · lgx + 1, then f (100)=

f(100)=f(1/100)·lg100+1
=2f(1/100)+1
=2[f(100)·lg(1/100)+1]+1
=2[f(100)·(-2)+1]+1
Let f (100) = t
Then t = 2 [(- 2) t + 1] + 1
The solution is t = 3 / 5
So f (100) = 3 / 5