Let $f(x)=x^{3} +5x +3$ Prove that $f(x)$ is uniformly continuous on any closed interval.

Question

Let $$f(x)=x^{3} +5x +3$$

Prove that $f(x)$ is uniformly continuous on any closed interval.

My Attempt :

Let $[a,b]$ any closed interval such that $a<b$

I need to prove that for every $\epsilon > 0 $ exists $\delta > 0 $ such that if $x,y \in [a,b] , |x-y|<\delta \implies |f(x)-f(y)|<\epsilon$

$|(x^{3} +5x +3)-(y^{3} +5y +3)| = |5(x-y)+(x^3 - y^3)|=|(x-y)(x^2 + xy +y^2 +5)|$

now I need to find when $(x^2 + xy +y^2 +5) \leq$ something with $a or b$

so what i did is :

$4a^2\leq (x+y)^2 \leq 4b^2 \rightarrow 4a^2-xy+5 \leq (x^2 + xy +y^2 +5) \leq 4b^2-xy+5$

my question is can I assume the $4b^2-xy+5 \leq 4b^2+5$ to continue the question ?

Answer 1

No, you cannot. That inequality may well be false.

But you can do this: take $M\in(0,\infty)$ such that $\lvert a\rvert,\lvert b\rvert\leqslant M$. Then, if $x,y\in[a,b]$,$$\lvert x^2+xy+y^2+y\rvert\leqslant M^2+M^2+M^2+5=3M^2+5.$$