Verify that f(x) = x^2 + 2x, x ≤ 1 and 3x, x > 1 is continuous at x =1.                         

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should remember that a function is continuous at a point if there exists limit of the function at the given point and this limit has a value equal to value of function at the given point such that:

`lim_(x->1,x<1) f(x) = lim_(x->1,x>1) f(x) = f(1)`

`lim_(x->1,x<1) f(x) = lim_(x->1,x<1)(x^2 + 2x)`

You need to substitute 1 for x in equation under limit such that:

`lim_(x->1,x<1) (x^2 + 2x)= 1 + 2 = 3`

`lim_(x->1,x>1) f(x) = lim_(x->1,x>1) 3x = 3*1 = 3`

Youneed to evaluate f(1), hence, you need to select the equation of the function whose domain is `(-oo,1]`  such that:

`f(x) = x^2 + 2x`

`f(1) = 1 + 2 = 3`

Since there exists limit of the function at `x = 1, lim_(x->1)f(x) = 3`  and the value of the limit coincides to the value of the function at x = 1, hence, the function is continuous at x = 1.

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