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.
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.
Can you guys also help me on this one?
x^5 + 4x^4 -16x -16 < 0
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
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
Thank you sooo much!
But how can I solve the 5th degree one algebraically?
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.
To solve the first inequality, we need to find the values of 'b' that satisfy the inequality:
(1/(2b+1)) + (1/(b+1)) > (8/15)
To begin, let's find the common denominator for the left side of the inequality. The common denominator is (2b+1)(b+1). We can multiply each term on the left side by this common denominator to simplify the inequality:
[(b+1) + (2b+1)] / [(2b+1)(b+1)] > (8/15)
Simplifying further:
[3b + 2] / [(2b+1)(b+1)] > (8/15)
Now, let's get rid of the fraction by multiplying both sides of the inequality by (2b+1)(b+1):
[3b + 2] * [(2b+1)(b+1)] / [(2b+1)(b+1)] > (8/15) * [(2b+1)(b+1)]
This simplifies to:
3b + 2 > (8/15) * [(2b+1)(b+1)]
Next, let's distribute and simplify the right side:
3b + 2 > (8/15) * (2b^2 + 3b + 1)
To further simplify, distribute the (8/15) across the terms:
3b + 2 > (16/15)b^2 + (8/15)b + (8/15)
At this point, we have a quadratic inequality. Let's move all terms to one side to set it to zero:
(16/15)b^2 + (8/15)b + (8/15) - 3b - 2 > 0
Combine like terms:
(16/15)b^2 + (23/15)b - (2/15) > 0
Now, we can use a number line to solve this quadratic inequality. First, we need to find the critical points by setting the expression equal to zero:
(16/15)b^2 + (23/15)b - (2/15) = 0
We can solve this equation by factoring, completing the square, or using the quadratic formula.
Once you find the critical points, plot them on a number line. Then, pick test values for each interval and plug them into the inequality expression to determine which intervals satisfy the inequality.
The solution to the inequality will be the intervals on the number line where the expression is greater than zero.
Similarly, you can follow the same steps to solve the second inequality:
(2/(x+1)) < 1 - (1/(x-1))
To solve, find the common denominator on the left side, multiply both sides to eliminate the fraction, simplify, set it to zero, plot the critical points on the number line, and test intervals to determine the solution.
Remember to always double-check your solution by plugging the values from the intervals into the original inequality expression to ensure they satisfy the inequality.
Finally, you can express the solutions using interval notation, which involves using brackets or parentheses to indicate open or closed intervals on the number line.