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????

(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.

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.

To answer your questions:

a) To determine the intervals on which the function is increasing or decreasing, we need to find the derivative of f(x) and analyze its sign. Let's find f'(x):

f(x) = x/(1+x)^2

To find f'(x), we can use the quotient rule. The quotient rule states that for two differentiable functions u(x) and v(x), the derivative of the quotient u(x)/v(x) is given by (u'(x)v(x) - u(x)v'(x))/(v(x))^2:

f'(x) = [(1+x)^2 * (1) - x * 2(1+x)] / ((1+x)^2)^2
= (1 + 2x + x^2 - 2x - 2x^2) / (1 + 2x + x^2)^2
= (1 - x - x^2) / (1 + 2x + x^2)^2

Next, since f'(x) is a rational function, we need to find the critical points where f'(x) = 0 or is undefined. However, in this case, f'(x) is not factorable and finding an exact solution is difficult. So, we can analyze the intervals by using test values to evaluate the sign of f'(x). We can choose test values on each interval (-∞, -1), (-1, 1), and (1, +∞).

Let's consider the interval (-∞, -1):
Choose x = -2:
f'(-2) = (1 - (-2) - (-2)^2) / (1 + 2(-2) + (-2)^2)^2
= -3 / (-3)^2
= -3/9
= -1/3

Since f'(-2) = -1/3 < 0, f(x) is decreasing on the interval (-∞, -1).

Similarly, you can choose test values in the remaining intervals to determine if f(x) is increasing or decreasing.

b) To find local min and max values of f(x), we need to find the critical points and endpoints of the interval. Critical points occur where f'(x) = 0 or is undefined.

In this case, f'(x) is not factorable, so finding exact critical points is difficult. However, we can find approximate values by using numerical methods or a graphing calculator.

For f(-1), you got an undefined result. It's important to note that f(-1) is not defined because the denominator of the original function f(x) becomes zero, making the function undefined at x = -1.

f(1) = 1/4. So, the point (1, 1/4) is a local minimum value of f(x).

c) To find the intervals of concavity and inflection points, we need to find the second derivative of f(x), which represents the concavity of the function.

Let's find f''(x):

To find f''(x), we differentiate f'(x):

f'(x) = (1 - x - x^2) / (1 + 2x + x^2)^2

Using the quotient rule, we can differentiate the above function to find f''(x):

f''(x) = [(1 + 2x + x^2)^2(-1 - 2x) - (1 - x - x^2)(2(1 + 2x + x^2)(2x + 2))]/((1 + 2x + x^2)^2)^2

Expanding and simplifying this equation would yield a long expression, similar to what you obtained. However, factoring the numerator or finding an exact solution may not be possible or practical in this case.

Instead, you can examine the concavity of the function by analyzing the sign of f''(x) using test values in each interval. Choose test values in each interval (-∞, -1), (-1, 1), and (1, +∞) to determine the concavity of f(x).

To find the inflection points where the concavity changes, look for the values of x where f''(x) changes sign.

By analyzing the sign of f''(x) using test values, you can determine the intervals of concavity and potential inflection points.