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Therefore I would choose
Just need to check the following, can anyone help?
I need to use the epsilon-delta definition of continuity to prove that
f(x) = x^2 - 4x
is continuuous at 3
This is what I have,
The domain of f is R(the set of real numbers)
Let epsilon>0 be given. We want to choose delta>0, in terms of epsilon, such that
|f(x)-f(3)|<epsilon, for all x with |x-3|< delta (1.0)
1. First we write
f(x)-f(3) = x^2 - 4x + 3 = (x-1)(x-3)
2. Next we obtain an upper bound for |x+3|, when x is near 3.
If |x-3|<or equal to 1, than x lies in the closed interval [1,4], so
|x+3|<or eqaul to|x|+3(Triangle Inequality)
<or equal to 4+3=7
3. Hence
|f(x)-f(3)|<or equal to 7 |x-3|, for |x-3|<or equal to 1.
So if |x-3|< delta, where 0<delta<or equal to 1, then
|f(x)-f(3)|,7delta
Now 7delta<or equal to epsilon if and only if delta < or equal to 1/7 epsilon. Thus if we choose delta = min {1,1/7epsilon}, then
|f(x)-f(3)| < 7delta<or equal to 7 times 1/7 epsilon = epsilon, for all x with |x-3|< delta,
which proves staement (1.0).
Thus f is continuous at the point 3.