# Show that solutions of `x+2^(x-1)+sin(x-1)=1` are bigger than 0 and less then 1?

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You should move the constant term 1 to the left side, hence, you will form the function `f(x) = x + 2^(x-1) + sin(x-1) - 1`

You need to prove that the function has at least one root in interval `[0,1], ` hence, the graph of the function intersects at least one time x axis if `x in [0,1].`

You need to evaluate `f(0)` and `f(1)` such that:

`f(0) =0 + 2^(0-1) + sin(0-1) - 1 => 1/2 - sin 1 - 1 = -1/2 - sin 1 < 0`

`f(1) = 1 + 2^0 + sin 0 - 1 => f(1) = 1 + 1 + 0 - 1 = 1 > 0`

Notice that `f(0)` is negative and `f(1)` is positive, hence, there exists a value `c in [0,1]` such that `f(c) = 0` .

**Hence, the given function `f(x) = x + 2^(x-1) + sin(x-1) - 1` has at least one solution in [0,1]. **