Here is the question: Use the definition of uniform continuity to prove $f(x)=x^2 + 2x - 5$ is uniformly continuous on $[0,3]$.
I need help understanding how this works, and solving these types of problems in general that specify intervals.
Thanks in advance!
$|f'(x)| = |2x+2| \leq 2|x|+2 \leq 2\cdot 3+2 = 8$. Thus by MVT: $|f(x)-f(y)| = |f'(c)(x-y)| \leq 8|x-y|$. Can you take it from here?.
Method $2$:
$|f(x)-f(y)| = |x^2+2x-5 - (y^2+2y-5)| = |x-y||x+y+2|\leq (|x|+|y|+|2|)|x-y| \leq (3+3+2)|x-y| = 8|x-y|$. And you can go from here as well.