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Math 006 -Hints

This is a continuation of the hints page.

Question about assignement 8

Question

For the functions below, find all of the following, and sketch the graph of the function:

(i) Domain
(ii) Intercepts
(iii) Asymptotes
(iv) Intervals of increase / decrease
(v) Local maximum and minimum
(vi) Convextity, points of inflection (not in case D)

A. (x² - 4)³
B. (x² - 4)^1/2
C. (x - 1)/(x + 1)²
D. 1/(x(x + 2))

Hints

I'll do a calcultion for a function similar to, but different from the first example. Note, this case does not envolve asymptotes - see quiz page for an examples.

A1 f(x) = (9x² - 1)⁵

(vi) For convexity and points of inflection, you need to find the second derivative.

(Sorry I used the word concavity in question, which doesn't mean anything)

The following is the rule for saying what is what:

f '' (x) > 0	Convex up
f '' (x) < 0	Convex down
f '' (x) = 0	inflection

Proceedure for finding convextity and inflection:

(1) Find f ''(x)

(2) Factor f ''(x) as much as possible

(3) Where f'' (x) = 0 are inflection points

(4) Make a table, which is divided into intervals between inflection points, and labled with each factor of f ''(x)

(5) Work out whether each factor is positive or negative on the intervals

(6) Use (5) to find whether the whole function is positive or negative on each interval; this tells you whether it is concave up or down.

(1) So, first we find the first derivative:

f '(x) = 5*(9x² - 1)⁴*(2*9x) = 90x(9x² - 1)⁴

Next for the second derivative:

f ''(x) = 90x*4*(9x² - 1)³*(2*9x) + 90(9x² - 1)⁴

(2) When you are trying to find where a function is postive or negative, it's important to factorize it as much as possible.

So, here's the second derivative factored more:

f ''(x) = 90(9x² - 1)³*(72x+ (9x² - 1)) f ''(x) = 90(9x² - 1)³(9x² + 72x - 1)

Can we factor this more? Well, remember about differences of squares, and we get:

f ''(x) = 90((3x - 1)(3x+1))³(9x² + 72x - 1) f ''(x) = 90(3x - 1)³(3x+1)³(9x² + 72x - 1)

What about the (9x² + 72x - 1)

This has roots

(-72 +/-(72^2 + 4*9*1))/(2*9) The roots are approximately 1/9 and -1/9

This means that approximately

(9x² + 72x - 1) = (9x - 1)(9x + 1)

So this means that the whole of f''(x) can be written as

f ''(x) = 90(3x - 1)³(3x + 1)³(9x - 1)(9x + 1)

We are interested in when f ''(x) is positive, negative or zero.

(3) Since we've factored f ''(x), we can see where it will be zero; this is at the roots of f ''(x). These points are the inflection points of the function:

The inflection points of f(x) = (9x² - 1)⁵ are at

x = 1/3, -1/3, -1/9 , 1/9

So, in order, these are -1/3, -1/9, 1/9, 1/3

The intervals these give you are

(-infinity, -1/3), (-1/3, -1/9), (-1/9, 1/9), (1/9, 1/3), (1/3, infinity)

Make a table:

	factors									f ''(x)
intervals	90*9	(3x - 1)	(3x - 1)	(3x - 1)	(3x + 1)	(3x + 1)	(3x + 1)	(9x - 1)	(9x + 1)
(-infinity, -1/3)
-1/3	----------------------------------------------------------
(-1/3,-1/9)
-1/9	----------------------------------------------------------
(-1/9, 1/9)
1/9	----------------------------------------------------------
(1/9, 1/3)
1/3	----------------------------------------------------------
(1/3, infinity)

I've divided the table by lines labled with the places where f ''(x) is zero. we have to fill in signs now.

It's useful to fill in the zeros first, on the lines, for reference, eg 3x - 1 = 0 when x = 1/3

	factors									f ''(x)
intervals	90*9	(3x - 1)	(3x - 1)	(3x - 1)	(3x + 1)	(3x + 1)	(3x + 1)	(9x - 1)	(x + 1/9)
(-infinity, -1/3)
-1/3					0	0	0
(-1/3, -1/9)
-1/9									0
(-1/9, 1/9)
1/9								0
(1/9, 1/3)
1/3		0	0	0
(1/3, infinity)

Then for each factor, it's positive then zero, then negative (note, this is because I have factors of the form (a*x + b), with a positive; if a is negative, it's the other way round, eg, 1-x is negative for x < 1, the 0 at x=1, then positive on x > 1).

For instance, 3x - 1 < 0 for x < 1/3, and 3x - 1 > 0 for x > 1/3, so we can fill in the 3x - 1 columns:

	factors									f ''(x)
intervals	90*9	(3x - 1)	(3x - 1)	(3x - 1)	(3x + 1)	(3x + 1)	(3x + 1)	(9x - 1)	(x + 1/9)
(-infinity, -1/3)		-	-	-
-1/3					0	0	0
(-1/3, -1/9)		-	-	-
-1/9									0
(-1/3, 1/9)		-	-	-
1/9								0
(1/9, 1/3)		-	-	-
1/3		0	0	0
(1/3, infinity)		+	+	+

Use the same kind of reasoning to fill in the rest of the columns under the factors:

	factors									f ''(x)
intervals	90*9	(3x - 1)	(3x - 1)	(3x - 1)	(3x + 1)	(3x + 1)	(3x + 1)	(9x - 1)	(x + 1/9)
(-infinity, -1/3)	+	-	-	-	-	-	-	-	-
-1/3					0	0	0
(-1/9, -1/3)	+	-	-	-	+	+	+	-	-
-1/9									0
(-1/3, 1/9)	+	-	-	-	+	+	+	-	+
1/9								0
(1/9, 1/3)	+	-	-	-	+	+	+	+	+
1/3		0	0	0
(1/3, infinity)	+	+	+	+	+	+	+	+	+

Finally multiply the signs to get the sign of f ''(x) at the end of each row:

	factors									f ''(x)
intervals	90*9	(3x - 1)	(3x - 1)	(3x - 1)	(3x + 1)	(3x + 1)	(3x + 1)	(9x - 1)	(x + 1/9)
(-infinity, -1/3)	+	-	-	-	-	-	-	-	-	+
-1/3					0	0	0			0
(-1/3, -1/9)	+	-	-	-	+	+	+	-	-	-
-1/9									0	0
(-1/9, 1/9)	+	-	-	-	+	+	+	-	+	+
1/9								0		0
(1/9, 1/3)	+	-	-	-	+	+	+	+	+	-
1/3		0	0	0						0
(1/3, infinity)	+	+	+	+	+	+	+	+	+	+

Note, usually, the zero line is not drawn in, so you get a table that looks like:

	factors									f ''(x)
intervals	90*9	(3x - 1)	(3x - 1)	(3x - 1)	(3x + 1)	(3x + 1)	(3x + 1)	(9x - 1)	(x + 1/9)
(-infinity, -1/3)	+	-	-	-	-	-	-	-	-	+
(-1/3, -1/9)	+	-	-	-	+	+	+	-	-	+
(-1/9, 1/9)	+	-	-	-	+	+	+	-	+	+
(1/9, 1/3)	+	-	-	-	+	+	+	+	+	-
(1/3, infinity)	+	+	+	+	+	+	+	+	+	+

But I think the zeros are useful.

So, we get

on (-infinity, -1/3)	f(x) is concave up
on (-1/3, -1/9)	f(x) is concave down
on (-1/9, 1/9)	f(x) is concave up
on (1/9, 1/3)	f(x) is concave down
on (1/3, infinity)	f(x) is concave up

(v),(iv) Intervals of increase and decrese, and local max and min.

This is very similar to the above, but instead of doing a table for f ''(x), we do a table for the sign of f'(x). All the same kind of things apply, so I'm not going to rewrite all the details.

f ' (x) > 0	increasing
f ' (x) < 0	decreasing
f ' (x) = 0	local max or min

f ' (x) goes +, 0 , - indicates a local max

f ' (x) goes -, 0 , + indicates a local min

f ' (x) goes -, 0 , - indicates a point of inflection

f ' (x) goes +, 0 , + indicates a point of inflection

Proceedure for finding intervals of increase, decrease, and local max and min:

(1) Find f '(x)

(2) Factor f '(x) as much as possible

(3) Where f' (x) = 0 are critical numbers

(4) Make a table, which is divided into intervals between critical numbers, and labled with each factor of f '(x)

(5) Work out whether each factor is positive or negative on the intervals

(6) Use (5) to find whether the whole of f '(x) is positive or negative on each interval; this tells you whether it is increasing or decreasing, and what kind of critical points you have

First we have:

f '(x) = 90x(9x² - 1)⁴ =90x(3x + 1)⁴(3x + 1)⁴

The roots of f '(x) are x = 0, x = 1/3, and x = -1/3, so these are the critical numbers.

So the table looks like this:

	factors										f '(x)
Interval	90	x	(3x + 1)	(3x + 1)	(3x + 1)	(3x + 1)	(3x - 1)	(3x - 1)	(3x - 1)	(3x - 1)
(-infinity, -1/3)	+	-	-	-	-	-	-	-	-	-	-
-1/3	+	-	0	0	0	0	-	-	-	-	0
(-1/3,0)	+	-	+	+	+	+	-	-	-	-	-
0	+	0	+	+	+	+	-	-	-	-	0
(0,1/3)	+	+	+	+	+	+	-	-	-	-	+
1/3	+	+	+	+	+	+	0	0	0	0	0
(1/3,infinity)	+	+	+	+	+	+	+	+	+	+	+

So, we get

on (-infinity, -1/3)	f(x) is decreasing
on (-1/3 ,0 )	f(x) is decreasing
on (0, 1/3)	f(x) is increasing
on (1/3, infinity)	f(x) is increasing

at -1/3 and 1/3 there is an inflection point

at 0 there is a local minimum

For the remaining parts of this question for example A1,

(i): the domain is all of R - this function is always defined

(ii): intercepts: if x = 0, then y = (9*0² - 1)⁵ = 1

If y = 0, then x = 1/3 or -1/3

(iii): Asymptotes - this function has no asymptotes

Finally, we can sketch the curve, paying attention to what happens at x = -1/3, 1/3, 0, 1/9, -1/9

At the local minium, the value of the function is -1.

We get something like the following:

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