Proof that a function is continuous

[LASmith]

Prove that the function is continuous when f(x)=0
f(x)=x⁴-7x³+11x²+7x-12f(c)-\epsilon<f(x)<f(c)+\epsilon

Limits maybe taken, however, we do not have the value for c in the limit equation.

 

[SammyS]

It looks like you have to solve for x, when f(x) = 0, i.e., find the zeros of f(x).

For one of the zeros, notice that the sum of the coefficients of f(x) is zero. Therefore, f(1) = 0.

So you know that one of the factors of f(x) is (x-1). Use long division or synthetic division to find g(x) such that: f(x) = (x-1)g(x).

Added in Edit.
Notice that: f(-x) = x⁴+7x³+11x²-7x-12. Therefore, f(-(1)) = 0 .

 

[LASmith]

It looks like you have to solve for x, when f(x) = 0, i.e., find the zeros of f(x).

Solving this gives (x-1)(x+1)(x-4)(x-3)
But is this sufficient to show that it is continuous?

 

[SammyS]

Solving this gives (x-1)(x+1)(x-4)(x-3)
But is this sufficient to show that it is continuous?

Of course not.

The problem is to show that f(x) is continuous when f(x)=0. So the problem has become: show that f(x) is continuous for x = -1, 1, 3, 4 .