# Thread: maximum value of f(x) for each positive x

1. ## maximum value of f(x) for each positive x

Can anybody guide me on this problem?

thank you.

2. ## Re: maximum value of f(x) for each positive x

Let y = x + 1/x. Consider $\displaystyle y^3$ and express $\displaystyle x^3 + 1/x^3$ through y. Then express $\displaystyle x^6 + 1/x^6$ through $\displaystyle (x^3 + 1/x^3)^2$ and thus through y.

3. ## Re: maximum value of f(x) for each positive x

you mean to say that i am going to substitute the numerator first quantity by y and so it is y^3 then the second quantity would be solely y?
then the denominator as well? pls guide me on this sir so confused to find what does maximum value

4. ## Re: maximum value of f(x) for each positive x

Originally Posted by rcs
you mean to say that i am going to substitute the numerator first quantity by y and so it is y^3 then the second quantity would be solely y?
then the denominator as well? pls guide me on this sir so confused to find what does maximum value
Substitute $\displaystyle y = x + \frac{1}{x}$ just like emakarov suggested. So $\displaystyle (x + \frac{1}{x})^3 = y^3$.

Also, $\displaystyle (x + \frac{1}{x})^3 = x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x})$. Substitute $\displaystyle x + \frac{1}{x}$ to obtain

$\displaystyle y^3 = x^3 + \frac{1}{x^3} + 3y \Rightarrow x^3 + \frac{1}{x^3} = y^3 - 3y$

Then you can substitute this into the original f(x).

5. ## Re: maximum value of f(x) for each positive x

thank you i got what it meant... but how about the denominator sir? do i need to substitute it also? do i need only to manipulate the x^6 + 1/x^6?

6. ## Re: maximum value of f(x) for each positive x

Yeah, you're gonna have some more algebraic manipulation. Hint: What is $\displaystyle (x^3 + \frac{1}{x^3})^2$?

7. ## Re: maximum value of f(x) for each positive x

little help pls still cant get the point

8. ## Re: maximum value of f(x) for each positive x

Originally Posted by richard1234
Yeah, you're gonna have some more algebraic manipulation. Hint: What is $\displaystyle (x^3 + \frac{1}{x^3})^2$?
i got it x^6 +1/x^6 + 2.... need little guide here sir... may i know the next step ?
thank u

9. ## Re: maximum value of f(x) for each positive x

Originally Posted by rcs
i got it x^6 +1/x^6 + 2.... need little guide here sir... may i know the next step ?
thank u
Substitute into the denominator..

10. ## Re: maximum value of f(x) for each positive x

i made the hint given by richard1234 (x^3 + 1/x^3) ^2

and so i have obtained this

f(x)= (y^3 + y^3 - 3y) / (y^6 - (y^3 - 3y)^2 - 2 - 2)

= (y^3 + y^3 - 3y) / (y^6 - ( y^6 + 6y^4 + 9y^2) - 4
= (y^3 + y^3 - 3y) / 6 y^4 - 9y^2 - 4
= y(2y^2 - 3) / 3y^2 ( 2y^2 - 3) - 4

im stuck here ...

11. ## Re: maximum value of f(x) for each positive x

Eh, that looks right, as long as your algebra's correct.

Now you have a function in terms of y. Maximize it.

12. ## Re: maximum value of f(x) for each positive x

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