Question

I have a question.

Answer 1

what is it.

Answer 2

Here’s my question... you see that *-3x^-2*..... i wanna bring that *x^-2* down in the denominator... right... is this what i need to do..wait hold up, let me draw.

Answer 3

But it don’t seem right...

Answer 4

\[2x + \frac{y}{x} \cancel{=} \frac{2x+y}{x}\]

Answer 5

so that looks good

Answer 6

Yes but that x^-2 is only multiplying with -3

Answer 7

yes. why would you divide all the term with x^2?

Answer 8

I just wanna make that x^-2 positive.. that’s all

Answer 9

So i gotta bring it down in the denominator

Answer 10

yes

Answer 11

If you want you could do
\[f'(x) = \frac{3x^{2}(x^{2})+2(x^{2})-3}{x^{2}}\]

Answer 12

\[z+ (x*y^{-1})= z+\frac{x}{y}\]

Answer 13

That’s what i did nnesha sister. But it looks uglee

Answer 14

doesn't matter if it looks ugly or pretty. mathmatically it should be "correct"

Answer 15

\[\frac{ 3x^2+2-3 }{ x^2 }\cancel{=} 3x^2+2-\frac{3}{x^2}\]

Answer 16

Yeah but i just had to include that... as an alternative solution.. but what Chen did...makes it final answer contain a lot of redundant x^2 but i see what he did... thanks for your valuable input guys. Really appreciated. God bless!

Answer 17

that's a totally different from the actual answer.
if you're comfortable doing extra work go with the way mchen did but i don't see a point of multiplying the first two term with x^2 bec after all you u have to cancel it out

Answer 18

Ok thanks guys ^~^ bless you!!

Answer 19

thanks o^_^o