Monotonic function

Question

Is the function $f:[-1,1] \to \mathbb{R}$, where $$f(x) = x^2+2x+2$$ a strictly increasing function or (non-strictly) increasing function?

I calculated $f'(x) = 2x+2$, so $$f'(x) = 0 \Rightarrow 2x+2 = 0 \Rightarrow x = -1 \\ \Rightarrow f'(x) \geq 0 \quad$$ when $x \in [-1,1]$.

It means that $f(x)$ is strictly increasing in $[-1,1]$? Or (non-strictly) increasing in $[-1,1]$?

Answer 1

$f'(x) >0$ for $x >-1$ so $f$ is strictly increasing in $(-1,1]$. But continuity of $f$ implies that it is actually strictly increasing in $[-1,1]$.

Answer 2

To show that $f$ is strictly increasing on $[-1,1]$, we need to show that for $a,b\in [-1,1]$ $$a < b \implies f(a) < f(b)$$ Thus, let $a,b\in [-1,1]$, with $a < b$.

Applying the Mean Value Theorem, it follows that $$f'(c)=\frac{f(b)-f(a)}{b-a}$$ for some $c\in(a,b)$.

From $c > a$ and $a\ge -1$, we get $c > -1$, hence \begin{align*} &f'(c) > 0\qquad\text{[since $f'(x) > 0$ for $x > -1$]}\\[6pt] \implies\;&\frac{f(b)-f(a)}{b-a} > 0\\[6pt] \implies\;&f(b)-f(a) > 0\\[6pt] \implies\;&f(a) < f(b)\\[6pt] \end{align*} therefore $f$ is strictly increasing on $[-1,1]$.

Answer 3

$y=(x+1)^2 +1;$

Let $a<b$, $a,b \in [-1,1]$, i.e $a +b >-2$.

$y(b)-y(a)=$

$ (b+1)^2-(a+1)^2=$

$ (b-a)(a+b+2) >0,$

and we are done.