Given the function f (x) = (x ^ 2-3x-2) g (x) + 3x-4, where g (x) is a function whose domain is r, it is proved that the equation f (x) = 0 must have real roots in (1,2)

If there is a problem with the title, it should be f (x) = (X & sup2; - 3x + 2) g (x) + 3x-4. In the preceding brackets, it should be + 2, not - 2
Certificate:
When G (x) includes 0, when G (x) = 0
f(x)=3x-4
3x-4=0
X = 4 / 3, between (1,2)
When G (x) does not include 0
f(x)=g(x)x²+3x[1-g(x)]+2g(x)-4
f(1)=g(x)+3-3g(x)+2g(x)-4=-10
F (x) continuous, f (1) 0
There must be an X on (1,2) such that f (x) = 0
In conclusion, the equation f (x) = 0 must have real roots in (1,2)

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