simple calculus?

Question

Is f (x)= x^2 -3x-4 continuous for all real valued x? why is that?

Answer 1

Yes. There is no value of x other than x -> infinity where f(x) -> infinity.

Answer 2

Everything everyone has said so far is true. Thought I'd just throw in that you could use the epsilon/delta definition to prove it as well. We want to show for every epsilon there is a delta such that:
|f(x) - f(x')| < epsilon whenever |x - x'| < delta

We want to show this for every real x'.
Then:
|f(x) - f(x')|
= |x^2 - 3x - 4 - x'^2 + 3x' + 4|
= |x^2 - x'^2 - 3(x - x')|
= |x - x'||x + x' - 3|
< delta*(|x + x'| + 3)
= delta*(|x -x' +2x'| + 3)
<= delta*(|x - x'| + 2|x'| + 3)
< delta*(delta + 2|x'| + 3)
So all we have to do is choose delta so that epsilon = delta*(delta + 2|x'| + 3). Then we would have what we originally wanted: |f(x) - f(x')| < epsilon whenever |x - x'| < delta.

Answer 3

Yes, all polynomials are continuous. If you want to prove this, you can simply take note of the fact that constant functions and the function f(x)=x are continuous, and that sums and products of continuous functions are continuous, so all integer powers of x are continuous (since they are just a product of several copies of x, which is a continuous function), so all monomials are continuous (since they are products of a constant and an integer power of x, both of which are continuous functions), so all polynomials are continuous (since they are just sums of monomials, which are continuous functions).

Answer 4

if x^2-3x-4=0

f(x) is continuous do to

b^2-4ac>0

x1=4 and x2=-1
-1 x<-1 andx>4 f(x) is plus