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)
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