# Thread: Could be easy, but I am stumped.

1. ## Could be easy, but I am stumped.

If $\displaystyle f(x)=3x^2-\frac{8}{x '}\$ , find the positive value of $\displaystyle x$ for which $\displaystyle f(x+1)-f(x)$ is least.

I don't think I have ever seen a function with x' mixed up in there and I can't seem to even get started. Thanks in advance for any help.

2. Originally Posted by jabroni1212
If $\displaystyle f(x)=3x^2-\frac{8}{x '}\$ , find the positive value of $\displaystyle x$ for which $\displaystyle f(x+1)-f(x)$ is least.

I don't think I have ever seen a function with x' mixed up in there and I can't seem to even get started. Thanks in advance for any help.
Are you sure that the prime on the x is not just a comma (not part of the formula at all)?

3. I just tried it without the x' just as x, and I got the right anwer. Thanks. Wierd it really looks like an x'.

4. Originally Posted by jabroni1212
I just tried it without the x' just as x, and it doesn't seem to minimize. I am pretty sure it is as I wrote above.
$\displaystyle 3x^2 - \frac{8}{x}$ has a (relative) minimum. You just have to work at it a bit.

Otherwise, I'd just treat the x' as a constant.

-Dan

5. Thats a disappointing answer, I had several theories, but was keeping my mouth shut until someone more educated than me found the correct answer. :P