f = x^2/(x^2+4)
f' = 8x/(x^2+4)^2
f" = -8(3x^2-4)/(x^2+4)^3
Now, we have
f" = 0 at x = ±2/√3
That means
f(x) is concave down on (-∞,-2/√3)U(2/√3,∞), where f" < 0
f(x) is concave up on (-2/√3,2/√3) where f" > 0
Consider the function below.
f(x)= (x^2)/((x^2+4)
Find the interval where the function is concave up. (Enter your answer using interval notation.)
Find the intervals where the function is concave down. (Enter your answers using interval notation.)
1 answer
1 answer