how would you calculate m for continuos function f(x)?

f(x)=x^2-2x+m, x=<1

f(x)=e^x-e, x>1

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You need to remember the functions' continuity condition such that:

`lim_(x-gta,xlta)f(x)=lim_(x-gta,xgta)f(x)=f(a)`

The problem provides the information that the function keeps its continuity at x=1, hence:

`lim_(x-gt1,xlt1)f(x)=lim_(x-gt1,xgt1)f(x)=f(1)`

You need to evaluate left limit, hence you need to substitute 1 for x in `x^2-2x+m` such that:

`lim_(x-gt1,xlt1)(x^2-2x+m)= 1 - 2 + m =gt lim_(x-gt1,xlt1)(x^2-2x+m)= m-1`

You need to evaluate right limit, hence you need to substitute 1 for x in `e^x - e` such that:

`lim_(x-gt1,xgt1)(e^x - e)=e - e = 0`

You need to set equations `lim_(x-gt1,xlt1)(x^2-2x+m)=m-1` and `lim_(x-gt1,xgt1)(e^x - e)=0` equal such that:

`m-1=0 =gt m=1`

**Hence, evaluating m under provided conditions yields m=1.**

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