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Finding the maximum of the function watch

1. I am given the following function:
f(x) = {(1-x)^(-2/3) + (1-4x)^(-1/3)}/{(1-3x)^(-1/3)+ (1-4x)^(-1/4)}
and I need to deduce that it has a maximum at x=0.

Things I tried:
1) I tried differentiating the function and setting it equal to zero but it did not work as I could not solve the equation
2) I used Taylor expansion around x=0 and obtained that f(x) = 1 - 7/36x^2 +...

2. (Original post by spacewalker)
I am given the following function:
f(x) = {(1-x)^(-2/3) + (1-4x)^(-1/3)}/{(1-3x)^(-1/3)+ (1-4x)^(-1/4)}
and I need to deduce that it has a maximum at x=0.

Things I tried:
1) I tried differentiating the function and setting it equal to zero but it did not work as I could not solve the equation
2) I used Taylor expansion around x=0 and obtained that f(x) = 1 - 7/36x^2 +...

Well your Taylor expansion fits the bill. For sufficiently small x, the "-7/36x^2" term will dominate, f(0)=1, and either side of x=0, f(x) < 1.

Note that with diferentiation, you don't have to solve the equation, only show that f'(0)= 0, and f''(0) < 0
3. (Original post by ghostwalker)
Well your Taylor expansion fits the bill. For sufficiently small x, the "-7/36x^2" term will dominate, f(0)=1, and either side of x=0, f(x) < 1.

Note that with diferentiation, you don't have to solve the equation, only show that f'(0)= 0, and f''(0) < 0
Thank you very much for your reply! The only thing I am not sure about is that when you find f''(x) you get -7/8 and there is no x in it to sub x=0 (i.e. f''(x) is independent from x). So, is the fact that -7/8 is negative be sufficient to prove that x=0 is maximum?
4. (Original post by spacewalker)
Thank you very much for your reply! The only thing I am not sure about is that when you find f''(x) you get -7/8 and there is no x in it to sub x=0 (i.e. f''(x) is independent from x). So, is the fact that -7/8 is negative be sufficient to prove that x=0 is maximum?
As long as you've found f'(0)=0

The second derivative being independent of x is not an issue, other than the fact that it isn't, so i suspect you've made a slip in your working.

The only way the second derivative can be a constant is if the original function was a quadratic.
5. (Original post by ghostwalker)
As long as you've found f'(0)=0

The second derivative being independent of x is not an issue, other than the fact that it isn't, so i suspect you've made a slip in your working.

The only way the second derivative can be a constant is if the original function was a quadratic.
Oh, the thing is I used the approximation to find that f'(0) = 0 and then I differentiated it again to find f''(x). Would not that be a suitable method?
6. (Original post by spacewalker)
Oh, the thing is I used the approximation to find that f'(0) = 0 and then I differentiated it again to find f''(x). Would not that be a suitable method?
Ok, that will work, but just be aware that you've only taken the Taylor series as far as the x^2 term (a quadratic), and ignored higher terms. If you hadn't stopped at the x^2 term when you take the second derivative you would have had higher order terms, it wouldn't be a constant. But because you're evaluating it at the point x=0, those terms vanish, and you can get away with it.

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