[SPLIT] First order critical numbers: f(x)= (x^2)/(x^2 + 3)

[rachelshelby10]

I have one more question regarding the same type of problem (finding first-order critical numbers)..... I just want to know if I got the answer right.......

f(x)= (x^2)/(x^2 + 3)

[(x^2 + 3)D(x^2)-(x^2)D(x^2 + 3)]/(x^2 + 3)^2

[(x^2 + 3)(2x)-(x^2)(2x)]/(x^2 + 3)^2

[2x^3 + 6x - 2x^3]/(x^2 + 3)^2

f '(x)= 6x.... is what I got for the derivative and

(0) is the only critical number I got by setting (6x=0)...is this right?

 

[stapel]

f(x)= (x^2)/(x^2 + 3)

[(x^2 + 3)D(x^2)-(x^2)D(x^2 + 3)]/(x^2 + 3)^2

[(x^2 + 3)(2x)-(x^2)(2x)]/(x^2 + 3)^2

You can use the fact that you have a common factor of "2x" in the numerator:

. . . . .\(\displaystyle \frac{2x(x^2\, +\, 3\, -\, x^2)}{(x^2\, +\, 3)^2}\)

. . . . .\(\displaystyle \frac{2x(3)}{(x^2\, +\, 3)^2}\)

But the result is the same: A fraction is zero when the numerator is zero, so f' = 0 for 6x = 0, or x = 0.

Good work!