(a) and (b) are correct
(c) what's the problem?
f"(x) = 2(x-2)/(x+1)^4
the denominator is positive always.
So,
f" < 0 for x < 2
f" > 0 for x > 2
Now you can say where f is concave up or down, and where the inflection point is.
f(x)= x/(1+x)^2
a) find the intervals on which f is increasing or decreasing.
Is it ....f is decreasing on the intervals (- infinity, -1) and (1, infinity)...f is increasing on the interval (-1, 1).......
b)find the local min and max values of f..
For f(-1) I got undefined...
f(1)=1/4.......
So is the answer... no min, max at (1,0.25)?????
I think that is what the graph shows...
c) find the intervals of concavity and the inflection points.
I am stuck here!!! For f"(x) I got a very long answer.. ????? I got (-6x^5+4x^4-4x^3-10x^2-14x-4)/(x+1)^8.... To find the intervals of concavity, I know that I have to set f"(x)=0 but I could not factor the polynomials!!! Is it possible to factor this polynomial out to solve for x?did I even get f"(x) correctly????
2 answers
You went to too much work on f"
f' = (1-x)/(1+x)^3
f" = [(-1)(x+1)^3 - 3(1-x)(x+1)^2]/(x+1)^6
= [-(x+1)+3(x-1)]/(x+1)^4
= (-x-1+3x-3)/(x+1)^4
= (2x-4)/(x+1)^4
Just because the quotient rule says you wind up with v^2 in the bottom, you don't have to leave it like that.
f' = (1-x)/(1+x)^3
f" = [(-1)(x+1)^3 - 3(1-x)(x+1)^2]/(x+1)^6
= [-(x+1)+3(x-1)]/(x+1)^4
= (-x-1+3x-3)/(x+1)^4
= (2x-4)/(x+1)^4
Just because the quotient rule says you wind up with v^2 in the bottom, you don't have to leave it like that.
1 answer