Who can find all intervals on which f(x)= x2 /x2+4 is decreasing?

Question

the 2 are square...( x square).i have a test soon but can not solve this. i would like to work for the problem if anybody can solve it.

Answer 1

f(x) = x^2 / (x^2 + 4)

To find the intervals of increase/decrease, you have to solve for the derivative f'(x) and then make it 0.

To solve for the derivative you can use the quotient rule.

f'(x) = [ (2x) (x^2 + 4) - (x^2)(2x) ] / [x^2 + 4]^2
f'(x) = [ 2x^3 + 8x - 2x^3 ] / [x^2 + 4]^2
f'(x) = 8x / [x^2 + 4]^2

Now, set f'(x) to 0.

0 = 8x / [x^2 + 4]^2

Critical values are what make f'(x) = 0 or what make f'(x) undefined. In our case, we don't have values that make this undefined (since x^2 + 4 = 0 has no real solutions), but we do have one case where f'(x) = 0, and that is when the numerator,
8x = 0. If 8x = 0, then x = 0.

Our critical value is x = 0. To find the intervals of increase/decrease, test a value less than 0, and test a value greater than 0, in your function f'(x).

f'(x) = 8x / [x^2 + 4]^2

Test -1: Then f'(-1) = 8(-1) / 5^2 = -8/25, which is negative. Therefore, f is decreasing on (-infinity, 0].

Test 1: Then f'(1) = 8(1) / 5^2 = 8/25, which is positive.
Therefore, f is increasing on [0, infinity).

If we wanted to solve for a relative min, all we have to do is solve for f(0).
f(0) = 0^2 / (0^2 + 4) = 0/(0 + 4) = 0.
Relative min at (0,0)

Answer 2

f(x) = x^2 / (x^2 + 4)
f'(x) = [2x (x^2 + 4) - x^2 (2x)] / (x^2 + 4)^2 by quotient rule
= 8x / (x^2 + 4)^2

Now the denominator is always strictly positive, so f'(x) will be negative exactly when x is negative. So f(x) is decreasing on (-infinity, 0] (and increasing on [0, infinity), for reference).

Answer 3

f(x) = x^2 / (x^2+4)
f'(x) = (2x(x^2+4) - x^2(2x)) / (x^2+4)^2
f'(x) = (2x^3 + 8x - 2x^3) / (x^2+4)^2
f'(x) = 8x / (x^2+4)^2
f'(x) < 0 when x < 0,
so the function is decreasing while x is less than 0 and increasing, or if x > 0 and decreasing

Answer 4

as x moves from negative infinity to 0

and when x moves from 0 to positive infinity

Answer 5

I got an idea, why don't you graph the function and find out yourself?

Answer 6

are you in calculus? just take the derivative and see where it is negative with a sign chart

Answer 7

that's easy you should study to get it right or should I say practice lol