How would you find the intervals on which f is increasing and decreasing, local maximums and minimums, the intervals of concavity, and inflection points for this equation: f(x)=x^2/x^2+3. This should be read as x squared over x squared plus 3.
2006-07-17 11:17:00 · 5 answers · asked by johnnyboy16978 1 in Science & Mathematics ➔ Mathematics
To the third person: x squared plus 3 is on bottom while x squared is on top. The two terms shouldn't cancel out...
2006-07-17 11:30:57 · update #1
f'(x) = [(x^2+3)(2x) - (x^2)(2x)]/(x^2 +3)^2
f'(x) = [2x^3 + 6x - 2x^3]/(x^2 +3)^2
f'(x) = 6x/(x^2 +3)^2
x^2 + 3 never equals zero, so your only critical point is
6x = 0 or x = 0.
Test points less than zero and greater than zero.
f'(-1) < 0 and f'(1) > 0
increases on (0, infinity)
decreases on (-infinity, 0)
x = 0 is a local min. The point is (0,0).
For concavity, you go through the same process but you use the second derivative instead.
2006-07-17 11:41:11 · answer #1 · answered by MsMath 7 · 0⤊ 0⤋
find the first derivative and then graph it, on the intervals where it s y - value is negative, the actual function is decreasing, on the intervals where your derivative has positive y-value, the original funtion is increasing. Then make the derivative equal to zero, and solve for x, t he x value shoud be your local maximum or minimum. Then find the second derivative, and graph it. On the intervals where your second derivative is positive, your original function is concave up. Where the second derivativeis negative, your original function is concave down. Points of inflection are found by making the second derivative equal to zero and solving for x .
2006-07-17 11:24:45 · answer #2 · answered by inDmood 3 · 0⤊ 0⤋
1. find the derivative of f, in this case you should use the quotient rule
f(x)= [(x^2+3)2x-x^2(2x) ]/(x^2+3)^2
2. find the intervals where f(x)>0, those are the intervals where the funcion is increasing, and
then find the intervals where f(x)<0, those are the intervals where f is decreasing.
3. the points where f(x) =0 are the critical points.
2006-07-17 11:25:08 · answer #3 · answered by lobis3 5 · 0⤊ 0⤋
f(x)= all real numbers; it increases by an interval of escalating odd numbers
x=1 interval is 3 between x=2
x=2 interval is 5 between x=3
x=3 interval is 7 between x=4
and so on
2006-07-17 11:35:47 · answer #4 · answered by creative_idea_thinker 2 · 0⤊ 0⤋
I think that you have made a mistake in writing this equation: the x^2/x^2 would cancel out to 1 to give f(x) = 4 which is trivial. Check your equation!
2006-07-17 11:27:20 · answer #5 · answered by Auriga 5 · 0⤊ 0⤋