Question

How do I solve this ?

`f(x)=sqrt(x^2+3x)`, x belongs (0;1)
Prove that f'(x) > 0 for every x belongs (0;1)

Answer 1

`f'(x)=(2x+3)/(2sqrt(x^2+3x))`. As both numerator denominator is positive for all `x in (0,1)`, it follows that f'(x) >0 for all `x in (0, 1)`.

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(0, 1)`.

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

It's easier to derivate our expression if we use exponents, like this:
`f(x)=sqrt(x^2+3x)=(x^2+3x)^(1/2)`
Let's define the variable `u(x)=x^2+3x`

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

Therefore
`f'(x)=f'(u)u'(x)=(2x+3)/(2sqrt(x^2+3x))`

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

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

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

Therefore it follows:
`lim(x->0)f'(x)=oo`

We can also look at it this way: Substitute `1/y=x` in f'(x) where `y>1`.
We can then write:
`f'(x)=g(y)=(2/y+3)/(2sqrt((1/y)^2+3/y))`
`=1/2(3y+2)/sqrt(3y+1)`

As `3y+2>3y+1` for all values of `y>1`
we have `f'(x)=g(y)=(3y+2)/(2sqrt(3y+1))`
`>(3y+1)/(2sqrt(3y+1))=1/2sqrt(3y+1) > 1 >0` for all values of `y>1`

As `x=1/y` this is fulfilled for `f'(x)` when `x in (0, 1)`