Study continuity of a function at a point Hi I am trying to solve an exercise from Calculus 1, I have tried several things but I am confused and would like to know if I am solving it wrong.

A function f:[0,3] $\longrightarrow$ R satisfies the following property:
$|f(x)-3x+4| \leq 2(x-1)^2$
Check if it's continuous at $x_{0} = 1$

As soon as I saw the exercise I thought of solving it using the definition of continuity so by doing something like this:
$\forall \epsilon>0, \hspace{0.2cm} \exists \delta > 0 $ such that if $x \in I(x_{0}, \delta)$ then $|f(x)-f(x_{0})| < \epsilon$ and so I would try to reduce that to this and get something like this, assuming that $f(x_{0}) = 3x - 4$ (the exercise doesn't say it, is this implicit?) :
$|f(x)-3x+4| \leq 2(x-1)^2 \Longrightarrow   \sqrt\frac{|f(x)-3x+4|}{2} \leq (x-1)< \epsilon$
and so for the function to be continous we need to have $\delta = 3\epsilon^2$
Is this correct?
 A: First let me comment on your solution and then I'll tell you how I would go about solving it.
You ask if you can assume that $f(x_0)=3x-4$ by which I think you mean something like $f(x_0)=3x_0-4$ for $x_0=1$, i.e. substituting, $f(1)=-1$. This is true for this function but you certainly can't assume it, you need to prove it. Substituting $x=1$ into the given condition gives
$|f(1)+1|\leq 0$
i.e. $f(1)+1=0$ i.e. $f(1)=-1$.
When you take the square root of $(x-1)^2$ you write this as $x-1$ which is not in general correct. Another thing is that you seem to solve the problem backwards: the first thing you wrote is the definition of what it would mean for the function to be continuous. That has to be the last line of your proof, not the first one, because that's what you're trying to prove, not that which is given. Your choice for $\delta$ also doesn't seem correct, but it's hard to judge exactly because you didn't write your argument from start to finish.
Either way, let's see if we can solve this not relying directly on the definition (proving using $\epsilon-\delta$ directly should not usually be your first approach).

We showed $f(1)=-1$ so for $f(x)$ to be continuous at $x_0=1$ we need to show that $\lim_{x\to 1}f(x)=-1$. By the given inequality
$$0\leq |f(x)-3x+4| \leq 2(x-1)^2$$
The LHS and RHS both tend to $0$ as $x\to 1$, so by the squeeze theorem
$$\lim_{x\to 1}|f(x)-3x+4| =0$$
By theorems which should be familiar (can you justify this?) this implies $\lim_{x\to 1} f(x)-3x+4=0$ and therefore $\lim_{x\to 1} f(x)=-1$, which is what was to be shown.
A: I tried to do your exercise and ended up with that solution that does use the definition of continuity in some sense but does not involve all these $\varepsilon$ and $\delta$ etc... It seems easier to manipulate to me.
First put $x=1$ in your inequality to get $|f(1)+1| \leq 0$ so $f(1)=-1$. At this point we don't know if that is useful or not but this is a good reflex to have to gather as much information as possible.
Then we look what happens around the point $x=1$. Let $h$ such as $x=1+h$, then. We are going to show that $|f(1+h)-f(1)| \xrightarrow[h \to 0]{} 0 $ i.e that f is continuous at point $x=1$. This is another way to formulate the continuity.
\begin{align}
|f(1+h)-f(1)| &= |f(1+h)+1| \\
&= |f(1+h)-3\times(1+h)+4+3h| \\
&\leq |f(1+h)-3\times(1+h)+4|+3|h| \quad \text{(triangular inequality)} \\
 &\leq 2(1+h-1)^2 + 3|h|\quad \text{(using inequality with x=1+h)} \\ 
&= 2h^2 + 3|h| \\
&\xrightarrow[h \to 0]{} 0 \\
\end{align}
Therefore f is continuous.
Of course there is possibly lots of other way to proove it...
