# How do I solve this ?

## $f \left(x\right) = \sqrt{{x}^{2} + 3 x}$, x belongs (0;1) Prove that f'(x) > 0 for every x belongs (0;1)

May 13, 2018

$f ' \left(x\right) = \frac{2 x + 3}{2 \sqrt{{x}^{2} + 3 x}}$. As both numerator denominator is positive for all $x \in \left(0 , 1\right)$, it follows that f'(x) >0 for all $x \in \left(0 , 1\right)$.

#### Explanation:

Let's first draw a graph:
graph{(x^2+3x)^(1/2) [-2.671, 3.488, -0.507, 2.574]}

The derivated value in each point represents the gradient, which we can see is positive in the whole interval $x \in \left(0 , 1\right)$.

One way of solving this algebraically would be to show that f'(x) is positive for all values of $x \in \left(0 , 1\right)$.

It's easier to derivate our expression if we use exponents, like this:
$f \left(x\right) = \sqrt{{x}^{2} + 3 x} = {\left({x}^{2} + 3 x\right)}^{\frac{1}{2}}$
Let's define the variable $u \left(x\right) = {x}^{2} + 3 x$

Accfording to the chain rule for derivation (https://en.wikipedia.org/wiki/Chain_rule) we can write this as
$f ' \left(x\right) = f ' \left(u\right) u ' \left(x\right)$
We have
$u = {x}^{2} + 3 x \to u ' \left(x\right) = 2 x + 3$
and as $f \left(u\right) = {u}^{\frac{1}{2}}$
-> f'(u)=(1/2)u^((1/2)-1)=1/(2u^(1/2))=1/(2sqrt(u)

Therefore
$f ' \left(x\right) = f ' \left(u\right) u ' \left(x\right) = \frac{2 x + 3}{2 \sqrt{{x}^{2} + 3 x}}$

As both numerator and denominator in $f ' \left(x\right)$ is positive when $x > 0$, the ratio will also be positive. Therefore $f ' \left(x\right) > 0$

If a more formal proof is required, we notice that the numerator stays in the interval (3, 5) as $x \in \left(0 , 1\right)$.

The denominator, on the other hand, goes from 4 when x=1 to 0 when x=0.

Therefore it follows:
$\lim \left(x \to 0\right) f ' \left(x\right) = \infty$

We can also look at it this way: Substitute $\frac{1}{y} = x$ in f'(x) where $y > 1$.
We can then write:
$f ' \left(x\right) = g \left(y\right) = \frac{\frac{2}{y} + 3}{2 \sqrt{{\left(\frac{1}{y}\right)}^{2} + \frac{3}{y}}}$
$= \frac{1}{2} \frac{3 y + 2}{\sqrt{3 y + 1}}$

As $3 y + 2 > 3 y + 1$ for all values of $y > 1$
we have $f ' \left(x\right) = g \left(y\right) = \frac{3 y + 2}{2 \sqrt{3 y + 1}}$
$> \frac{3 y + 1}{2 \sqrt{3 y + 1}} = \frac{1}{2} \sqrt{3 y + 1} > 1 > 0$ for all values of $y > 1$

As $x = \frac{1}{y}$ this is fulfilled for $f ' \left(x\right)$ when $x \in \left(0 , 1\right)$