How do you prove that the function #f(x) = x^2 -3x +5# is continuous at a =2?

Answer 1

Prove using limit definition of continuity...

Any polynomial function is continuous everywhere, but let's prove this particular example using the limit definition of continuity...

A function #f(x)# is continuous at a point #a# if both #f(a)# and #lim_(x->a) f(x)# are defined and equal.

In our example:

#f(2) = 2^2-3(2)+5 = 4-6+5 = 3#
#f(2+h) = (2+h)^2-3(2+h)+5#
#= 2^2+4h+h^2-3(2)-3h+5#
#= 4-6+5+h+h^2#
#= 3+h+h^2 -> 3# as #h->0#
So #lim_(x->2) f(x) = lim_(h->0) f(2+h) = 3 = f(2)#
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Answer 2

To prove that the function f(x) = x^2 - 3x + 5 is continuous at a = 2, we need to show that the limit of f(x) as x approaches 2 is equal to f(2).

First, let's find the limit of f(x) as x approaches 2.

lim(x→2) (x^2 - 3x + 5) = (2^2 - 3(2) + 5) = (4 - 6 + 5) = 3

Next, let's find f(2) by substituting x = 2 into the function.

f(2) = (2^2 - 3(2) + 5) = (4 - 6 + 5) = 3

Since the limit of f(x) as x approaches 2 is equal to f(2), we can conclude that the function f(x) = x^2 - 3x + 5 is continuous at a = 2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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