Question

Solve each inequality:

(1/(2b+1)) + (1/(b+1)) > (8/15)

(2/(x+1)) < 1 - (1/(x-1))

My teacher told me the answer is in interval notation, and that a number line is used to solve.

Answer 1

let's sketch
y = 1/(2x+1) + 1/(x+1) and y = 8/5

http://www.wolframalpha.com/input/?i=plot+y+%3D+1%2F%282x%2B1%29+%2B+1%2F%28x%2B1%29+%2C+y+%3D+8%2F15

From the graph I guess the following:
-1 < x < -.6 OR -1/2 < x < 2
Notice that there are asymptotes at x = -1 and x = -1/2

How about algebraically
consider (1/(2b+1)) + (1/(b+1)) = (8/15)
multiply each term by 15(2b+1)(b+1)
15(b+1) + 15(2b+1) = 8(2b+1)(b+1)
45b + 30 = 16b^2 + 24b + 8
16b^2 - 21b - 22 = 0
(b-2)(16b+11) = 0
b - 2 or b = -11/16

Hey my guess was pretty close:

-1 < x < -11/16 OR -1/2 < x < 2

Put this in whatever "interval notation" you have been taught,
I only ever use the one shown above.

Answer 2

Can you guys also help me on this one?

x^5 + 4x^4 -16x -16 < 0

Answer 3

I will do this one without graphing it ....
(2/(x+1)) < 1 - (1/(x-1))
notice that x ≠ -1, +1

consider 2/(x+1) = 1 - (1/(x-1)
multiply each term by (x-1)(x+1)

2(x-1) = (x+1)(x-1) - (x+1)
2x - 2 = x^2 -1 - x - 1
x^2 -3x = 0
x(x-3) = 0
x = 0 or x = 3
so from our critical values of 0 and 3
we have 3 sections to consider

1. x < 0
2. 0 < x < 3
3. x > 3

pick any arbitrary values in each of those regions and sub it into the original .
Notice we don't really need the exact value, we just have to determine if the statement is true or false

1. le x = -5
LS = 2/-4 = -1/2
RS = 1 - 1/-6 = 7/6
is -1/2 < 7/6 ? YES, so it must be true for x < 0

2. let x = 2
LS = 2/3
RS = 1 - 1/1 = 0
is 2/3 < 0 ? NO

3. let x = 5
LS = 2/6 = 1/3
RS = 1 - 1/4 = 3/4
is 1/3 < 3/4, YES

so x<0 OR x > 3

Answer 4

x^5 + 4x^4 -16x -16 < 0
again how about looking at Wolfram
let x^5 + 4x^4 -16x -16 = 0

http://www.wolframalpha.com/input/?i=solve+x%5E5+%2B+4x%5E4+-16x+-16+%3D+0+

Nasty nasty!!!

Unfortunately there are no nice roots (x-intercepts)
There are 3 real intercepts and 2 complex roots
So where is the graph below the x-axis (the < 0 part ) ?
It looks like
x < - 4.17 OR -.88 < x < 1.66

Answer 5

Thank you sooo much!
But how can I solve the 5th degree one algebraically?

Answer 6

In general, there is no way to solve anything of 5th degree or higher. So, unless you can pick out some rational roots or easy quadratic factors, you are left with graphical or numeric methods.