If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

022. `query 22

Question: `qQuery problem 1 c    7th edition 3.1.4 derivative of -3x `sqrt(x+1) at x=-1, -2/3, 2`sqrt(3)/3 **** What is the derivative of the function at each of the indicated points?

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`a f(x) = -3x(x + 1)^(1/2) . Using the Product Rule we get

f ' (x) = (-3) ( x + 1)^(1/2) + 1/2(x +1)^(-1/2)(1)(-3x)

= -3(x+1)^(1/2) - 3/2 * x (x+1)^(-1/2).

You could substitute into this form and get the right answers. However it's good to see how to simplify this expression:

Simplifying, we multiply the first term by (x+1)^(1/2) / (x+1)^(1/2) to obtain

f ' (x) = -3(x+1) / (x+1)^(1/2) - (3 x / 2)/(x+1)^(1/2)

= [-3(x+1) - 3 x / 2 ] / (x+1)^(1/2)

= (-3x - 3 - 3 x / 2) / (x+1)^(1/2)

= (-6x - 6 - 3x) / ( 2(x+1)^(-1/2) )

= -3 ( 3x + 2) / ( 2(x+1)^(-1/2) )

Thus

f ' (x) = -3(3x + 2) / (2(x +1)^(1/2))

f ' (-1) is undefined (denominator is zero)

f ' (0) = -6/2 = -3

f ' (-2/3) = 0 (critical number, numerator is zero)

DER

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Question: `qQuery 3.1.8 increasing, decreasing behavior of x^2 / (x+1)

On what open intervals is x^2 / (x+1) increasing, and on what intervals is it decreasing?

Your solution:

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Given Solution:

`a We first look at where the function itself is positive, where it is negative, and where it is undefined:

The numerator of the function f(x) is always positive, being a square.

The denominator of f(x) goes from positive to negative at x = 1 and stays there, so the function is negative on the first two intervals, positive on the other two.

To determine where the function is increasing and where decreasing we need to find the derivative and determine where it is positive and where it is negative.

Using the quotient rule we get

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

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

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

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

0 = (x(x + 2)) / (x + 1)^2 .

The derivative is zero at x = 0 and x = -2, and undefined at x = -1.

So the function can change sign only at -2, -1 or 0.

The function therefore has the same sign on each of the intervals (-infinity, -2), (-2, -1), (-1, 0), (0 infinity).

The derivative goes from positive to negative at -2 and from negative to positive at 0. So the function is decreasing on (-2, 0) and increasing otherwise.

So the function is negative and increasing on (-infinity, -2), then negative and decreasing on (-2, -1), where it decreases asymptotic to the line x = 1.

Then the function is positive and decreasing on (-1,0), and positive and increasing on (0, infinity).

Note that x = -2 and x = 0 are extrema, with the function reaching a relative maximum of -4 at x = -2 and a relative minimum of 0 at x = 0. **

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`a** Query problem 3e    7th edition 3.1.30 critical numbers, increasing and decreasing intervals for f(x) = x / (x+1)

What are the critical numbers for the given function, and on what intervals is it increasing, and on what intervals is it decreasing?

Your solution:

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Given Solution:

`a Critical numbers are the numbers where the derivative is zero, and the points where the function is discontinuous.

f( x ) = x / (x +1) . Using the quotient rule we get:

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

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

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

We can set this expression equal to zero:

1 / (x+1)^2 = 0.

However we will get no solutions, since the numerator is never zero.

The only possible critical numbers are therefore those where the function is discontinous.

This occurs only at x = -1, which makes the denominator of f(x) = x / (x + 1) zero. The function is discontinuous at x = -1.

So x = -1 is a critical number.

The derivative, being the reciprocal of a perfect square, is positive at numbers less than -1 and is positive at numbers greater than -1. The derivative in undefined, and approaches infinity, as x -> -1.

On the interval (-infinity, -1), the derivative is positive and the function is therefore increasing. On the interval (-1, infinity), the derivative is positive and the function is therefore again increasing. **

A LITTLE MORE ABOUT THE GRAPH OF f(x):

You don't yet need to know the following, since asymptotes are covered in subsequent sections.  However if this makes sense to you, it's worth a look now while you are thinking about the function f(x) = x / (x + 1).

When x < -1, f(x) = x / (x + 1) has a negative numerator and a negative denominator.  f(x) is therefore positive when x < -1.

As it approaches x = -1 from the left, the graph approaches a vertical asymptote with the line x = -1, approaching the asymptote through positive values.

When -1 < x < 0, x is negative and x + 1 is positive, so f(x) = x / (x + 1) is negative.  When x = 0 is should be clear that f(x) = 0.  The graph has a vertical asymptote just to the right of x = -1, and it rises from negative values toward the point (0, 0). 

When x > 0, both x and x + 1 are positive, so the function takes positive values.  As x gets larger and larger the difference between numerator x and the denominator x + 1 becomes less and less significant and x / (x = 1) approaches 1.  So after passing through (0, 0) the graph continues to rise, but more and more slowly, becoming asymptotic to the horizontal line y = 1.

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