Proving that a function is an increasing function Question:
"5. Functions f and g, with domains $\mathbb{R}^{+}$, are defined as follows:
$$\text{f}:x \to \sqrt{x}, \quad \text{g}:x \to 1 + 3x^{2}.$$
If the function h is defined by $h(x) = f(g(x))$, show that h is an increasing function and sketch its graph."
My approach was to find h:
$\text{h}(x) = \sqrt{1 + 3x^{2}}$
Then find $\text{h}'(x)$:
$\text{h}'(x) = \dfrac{1}{2\sqrt{1 + 3x^{2}}} \cdot 6x = \dfrac{3x}{\sqrt{1 + 3x^{2}}}$
And then what I want to do is prove that $\text{h}'(x) > 0,\ \forall{}x \in \mathbb{R}^{+}$; but I'm, uh, not really sure where to begin?
 A: Since $x \in \mathbb R^+$, we have $x > 0$ and it is easy to see that $3x > 0$.
Also, note that $x > 0 \implies 3x^2 > 0 \implies 1 + 3x^2 > 0 \implies \sqrt{1+3x^2} > 0$.
Then, both numerator and denominator are positive, and you can draw your conclusion.
A: The derivative is alright, now set $\frac{3x}{\sqrt{1+3x^2}} \ge 0$. Note that the square root is always positive, so this comes down to $3x\ge 0 \implies x \ge 0$
Note: If you want to show that the function is strictly increasing then you should substitute $\ge$ with >
A: So as you have found, $h'(x)=\frac{3x}{\sqrt{1 + 3x^{2}}}$.
Now $x\in\Bbb{R}^+\Leftrightarrow x>0$.
The numerator, $3x$, is of the same sign as $x$, hence it's positive.
The denominator is a square root, and thus most be positive. As a result $h'(x)>0$ for $x>0$ and $h$ is increasing.
A: $3x \over \sqrt{3x^2+1}$ $> 0$ iff 3x > 0 since the denominator is always positive
Hence, h'(x) > 0 iff x > 0.
Since it is given that x > 0, it follows that h'(x) > 0.
A: Let $f$ and $g$ be increasing functions $\mathbb{R}\to \mathbb{R}$ show that the composite of the two functions $f.g $ wherever it is defined is increasing. Then show that $x \to \sqrt{x}$ and $x \to 1+3x^{2}$ are both increasing.  
Also , To show that $f(x)=3x/ \sqrt{1+3x^{2}}$ is increasing , divide both the numerator and denominator by $x$ the result is $3/ \sqrt{3+1/x^{2}}$ which is always positive .
