# How do I solve this ?

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#f(x)=sqrt(x^2+3x)# , x belongs (0;1)

Prove that f'(x) > 0 for every x belongs (0;1)

Prove that f'(x) > 0 for every x belongs (0;1)

##### 1 Answer

#### 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

One way of solving this algebraically would be to show that f'(x) is positive for all values of

It's easier to derivate our expression if we use exponents, like this:

Let's define the variable

Accfording to the chain rule for derivation (https://en.wikipedia.org/wiki/Chain_rule) we can write this as

We have

and as

Therefore

As both numerator and denominator in

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

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

Therefore it follows:

We can also look at it this way: Substitute

We can then write:

As

we have

As