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)

1 Answer
May 13, 2018

#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)#