Expert: Scott A Wilson - 11/28/2009

Question
Hi Scott,
I am studying for a calculus exam and I would like to get some advice on how to address each of the following questions regarding my practice problems:

(i) The domain of the function. (If not specified, assume the domain is the largestpossible on which the function makes sense.)
(ii) Horizontal and vertical asymptotes, if any.
(iii) Regions on which the function is increasing and on which the function is decreasing.
(iv) Local and global maxima and minima, if any.
(v) concavity and inflection points.

4) f(x) = 4x / (x^2+1).

I have found the derivative using the quotient rule.(4/(x^2+1)) - (8x^2 /(x^2+1)^2)
I set this equal to zero and solved.  x = + - squ rt of 1
I figured the domain to be set of all real numbers. (not completely sure as to how or why)
I think I need the second derivative to find inflection points and
concavity.
I wanted to approach this problem in the order it was laid out, but, I really am not sure exactly how to begin in. Can you help provide the proper steps/approach?

Thanks


Answer
(i) The domain of the function is whatever x values can be put in.
Since the denominator involves an x², an x² is never negative, and then 1 more is added,
the denominator is always positive.  Therefore the domain is all x.

(ii) The function is defined over all x, so there are no vertical asymptotes.
When x goes off to ±the function goes to 0, so y=0 is a horizontal asymptote.

(iii) Since the derivative of f(x) = g(x)/h(x) is (hg’ – gh’)/h², we find what g’ and h’ are.
Note that g and h are still g(x) and h(x).

It can be seen the g(x) =4x, so g’(x)=4.
It can be seen that h(x) = x²+1, so h’(x)=2x.
Thus, the derivative of f(x)=4x/(x²+1) is is ((x²+1)4 – 4x(2x))/(x²+1)².
The numerator turns out to be 4x²+4-8x² = -4x²+4 = -4(x²-1).
So f’(x) = -4(x²-1)/(x²+1)².  This can be seen to be 0 when x=±1.

When x is in the interval (-1,1), f’(x) is positive.  Outside this region, it is negative.

(iv) This means the local min is at –1 and the local max is at 1.
The value at –1 is 4(-1)/2 = -2 and the value at 1 is 4(1)/2 = 2.

(v) The concavity and inflection points are determined by the 2nd derivative.
That is, given f’(x) = -4(x²-1)/(x²+1)², find f”(x).
We have new values for g(x) and h(x) for f’(x)=g(x)/h(x).
The –4 will be left out front as a constant.
It is seen that g(x)=x²-1, so g’(x)=2x.  It is also seen that h(x)=(x²+1)²,
so h’(x)=2(x²+1)2x = 4x(x²+1).

The 2nd derivative is then f”(x) = 4(hg’-gh’)/h².
That is, f”(x) = 4((x²+1)²(2x) – (x²-1)4x(x²+1))/(x²+1)^4.
This can have every term divided by (x²+1), giving
f”(x) = 4((x²+1)(2x) – (x²-1)4x)/(x²+1)³.
Multiplying out the inside of the parenthesis in the numerator gives 2x³+2x – 4x³+4x.
This is the same as 6x³+6x=6x(x²+1).

Putting this back into f”(x) gives us, since 4*6=24, 24x(x²+1)/(x²+1)³.
Canceling an (x²+1) gives 24x/(x²+1)².
The denominator is always positive, so all we have to worry about is 24x.
That is positive when x is positive and negative when x is negative,
so the function is concave up when x is negative and concave down when x is positive.
There is an inflection point at x=0.

A graph of the function will be attached.