Is f(x) Increasing for Positive x and a in (0,1)?

[lys111]

I am supposed to show f(x) is increasing in x when x is real and positive.

f(x)= [(x+2)*(1-a^(x+1)] / [1-(x+2)*(1-a)*a^(x+1)-a^(x+2)]

a is any real in (0,1); x is real and positive

I have taken and first derivative of f(x):

f'(x)=1-a^(x+1)-a^(x+2)+(x+1)*(x+2)*log(a)*(a^(x+1)-a^(x+2))+a^(2x+3)

The problem is I cannot compare log(a) with the power of a. Can any of you genius help me with a proof as to showing f'(x) >0? Or maybe there is some other way to show f(x) is increasing in positive x?

 

[haruspex]

f(x)= [(x+2)*(1-a^(x+1)] / [1-(x+2)*(1-a)*a^(x+1)-a^(x+2)]

Please clarify the bracketing, preferably with LaTex. Is it
$f(x)=\frac{(x+2)*(1-a^{x+1})}{1-(x+2)*(1-a)*a^{x+1}-a^{x+2}}$ ?

 

[lys111]

Yes, it is. I am sorry about the typing/.

f(x)=\frac{(x+2)*(1-a^{x+1})}{1-(x+2)*(1-a)*a^{x+1}-a^{x+2}}