Increasing and decreasing functions

[Tanishq Nandan]

Homework Statement

f(x)=x[ax-x^2]^ (1/2) for a>0
Then,f(x)
A)increases on (3a/4 , a)
B)decreases on (0, 3a/4)
C)both A,B
D)None of these

Homework Equations

differentiation chain rule
f(x) is said to be increasing in (a,b) if it's derivative is positive and decreasing if it's derivative is negative for all x b/w a and b

The Attempt at a Solution

First af all,I found the domain of the given function which came out to be [0,a]
Now,the derivative of the function is:
(3ax-4x^2)/ [(4ax-4x^2)^1/2]
Now,the term in the denominator being inside square root is always positive,so we only need to deal with the numerator.
Which is:
x(3a-4x)
Now,due to it's domain x is also positive
Therefore the first term of the numerator is also positive,so it all comes down to the second term..
(3a-4x) which is positive(and hence the function increasing) for x b/w 0 and 3a/4 ,and negative for the rest.So,the corresponding option comes out to be D.
But,the answer given is C.
If anybody can point out where I am going wrong (or if the answer given is wrong,whichever),it qould be very helpful..

 

[Orodruin]

The answer given is wrong. Since the function is positive and goes to zero at x = 0 and x = a, it must increase in the beginning an decrease in the end. Here is a plot for a = 1:

Edit: If you don't want to bother differentiating the square root, you can also note that the function is positive in the domain and therefore is increasing/decreasing if its square is. It is much more convenient to differentiate $ax^3 - x^4$.

 

[Tanishq Nandan]

Got it