Question

prove that f(x)=\[\sqrt(x^2+x)\] is continuous on [0,infty)

Answer 1

how do u prove that something is continuous?

Answer 2

basically you have to prove that f(x) is defined on all the points from [0, inf), which is true.

Answer 3

wait are we talking about epsilon delta proof here?

Answer 4

ok i dont remember how to do this.

Answer 5

I think i remember now:
If you're in calc1, you gotta prove:
a) f(a) is defined
b) lim x->a [f(x)] exists
c) lim x->a [f(x)] = f(a)
for all a on [0.inf)

Answer 6

composition of continuous functions, thus.... or do we need e,d proof? @lordng93

Answer 7

i think @zzr0ck3r remembers this better than me.

Answer 8

It was last term, but .... lol. If they want a real proof ill write one up, but not sure what they need.

Answer 9

differentiability implies continuity as well.

Answer 10

\[f(x) is defined if x ^{2}+x \ge 0\]
\[x ^{2}+x+\frac{ 1 }{ 4 }\ge 0+\frac{ 1 }{ 4 }\]
\[\left| x+\frac{ 1 }{2 } \right|\ge \frac{ 1 }{ 2 }\]
\[x+\frac{ 1 }{ 2 }\le \frac{ -1 }{ 2 },x \le -1 and x+\frac{ 1 }{ 2 }\ge \frac{ 1 }{ 2 },x \ge0\]
\[ f(x) is defined \in (-\infty ,-1]\cup [0,\infty)\]
\[f'(x)=\frac{ 1 }{ 2\sqrt{x ^{2} +x} }\left( 2x+1 \right)\]
it exists in the domain except x=0
except 0 it is continuous in the domain.

Answer 11

thanks all and sorry for the late reply, my internet service is down until now