f = x^2/(x-4)^2
f' = -8x/(x-4)^3
f" = 16(x+2)/(x-4)^4
f is concave up where f" > 0, or x > -2
concave down then, is everywhere else, or x < -2
Unfortunately, there is an asymptote at x=4, so f is concave up on (-2,4)U(4,∞). The graph is concave up on both sides of the asymptote.
The graph is at
http://www.wolframalpha.com/input/?i=x^2%2F%28x-4%29^2
Consider the function below.
f(x)= (x^2)/((x-4)^2)
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.)
For interval where the function is concave up i got (-2,4) but the homework system keeps saying its wrong
and for intervals where the function is concave down i got (-oo,-2)U(4,oo) but homework system keeps saying its wrong
1 answer
1 answer