If $b>a>0$ show that $f(b)>f(a)$

Question

You have a function $f(a)=1+a^2$ where $f: \mathbb{R} \to \mathbb{R}$

If $b>a>0$ how would I show that $f(b)>f(a)?$

Answer 1

Just compute it:

$$f(b) - f(a) = (1 + b^2) - (1 + a^2) = b^2 - a^2 > 0$$

whenever $b > a$.

Answer 2

We have $$f'(x)=2x>0,\,\forall x>0$$ so $f$ is strictly increasing on the interval $(0,+\infty)$ hence if $b>a>0$ then $f(b)>f(a)$.

Answer 3

By plugging into your equation, of course!

Since $b > a > 0$, we have $b^{2} > a^{2}$. Thus, $f(b) - f(a) = (1 + b^{2}) - (1+a^{2}) = b^{2} - a^{2} > 0$, so $f(b) > f(a)$.