Inequalities

[S_100]

Solve − 1 < − (1/x) + 2x < 1.

The method I chose initally was to mutliply the whole inequality by x only to realise x could be negative and so multiplied out by x^2.

-x²<-x+ 2x³< x²
0 < 2x³ + x² - x < 2x²

I realised I couldn't do anything useful with the 2x² as substracting it would give an inequality, where solutions for x could not be found ( please correct me if wrong)

I then thought to sketch f(x) = 2x³ + x² - x , finding roots and y intercept,

and 2x², aswell, finding where both functions instersect

and find the area where y values of f(x) are greater than zero but less than the 2x² graph, and thus find the solutions for the corresponding x values.

My Question : Is there a quicker/easier method for finding solutions of x?

 

[Dr.Peterson]

The standard approach is to split the compound inequality into two, and solve each by getting 0 on one side, finding points where the other side can change sign (0 or undefined), and use that to find where it has the right sign. Then you will have to intersect the solutions to the two parts, namely −1 < −(1/x) + 2x and −(1/x) + 2x < 1.

Your method can be continued in a similar way, splitting into two separate quadratic inequalities and solving each in terms of signs.

 

[Steven G]

Solve − 1 < − (1/x) + 2x < 1.

The method I chose initally was to mutliply the whole inequality by x only to realise x could be negative and so multiplied out by x^2.

-x²<-x+ 2x³< x²
0 < 2x³ + x² - x < 2x²

I realised I couldn't do anything useful with the 2x² as substracting it would give an inequality, where solutions for x could not be found ( please correct me if wrong)

I then thought to sketch f(x) = 2x³ + x² - x , finding roots and y intercept,

and 2x², aswell, finding where both functions instersect

and find the area where y values of f(x) are greater than zero but less than the 2x² graph, and thus find the solutions for the corresponding x values.

My Question : Is there a quicker/easier method for finding solutions of x?

You can multiply by x Or other unknown quantities but you need to do 2 cases. Just understand that this may not always be helpful.

Case 1: Suppose x>0. Multiply by x but do NOT change the direction of the inequality. You need to be careful when you get your results and ONLY keep the part where x>0

Case 2: Suppose x<0. Multiply by x but DO change the direction of the inequality. You need to be careful when you get your results and ONLY keep the part where x<0

 

[S_100]

You can multiply by x Or other unknown quantities but you need to do 2 cases. Just understand that this may not always be helpful.

Case 1: Suppose x>0. Multiply by x but do NOT change the direction of the inequality. You need to be careful when you get your results and ONLY keep the part where x>0

Case 2: Suppose x<0. Multiply by x but DO change the direction of the inequality. You need to be careful when you get your results and ONLY keep the part where x<0

So would this be helpful as there could be a proof by contradiction for either conjecture to deduce whether x is actually positive or negative?

Say −1 < 3x + 4/ x − 6 < 1
suppose (x-6) > 0 and suppose ( x − 6 )<0
The (x-6) > 0 cannot be true by contradiction --> x> 6
3x + 4< x − 6 , which is incorrect
so we know x-6 has to be -ve.

 

[Dr.Peterson]

Say −1 < 3x + 4/ x − 6 < 1
suppose (x-6) > 0 and suppose ( x − 6 )<0
The (x-6) > 0 cannot be true by contradiction --> x> 6
3x + 4< x − 6 , which is incorrect
so we know x-6 has to be -ve.

Why would you suppose a contradiction?? That's nonsense.

Please, though, learn to use parentheses to say what you mean. I'm guessing you mean −1 < (3x + 4)/(x − 6) < 1, that is, [MATH]-1\lt \frac{3x+4}{x-6}\lt 1[/MATH], which is entirely different from what you wrote, which is [MATH]−1 < 3x + \frac{4}{x} − 6 < 1[/MATH] .

Have you learned the standard way to solve inequalities like [MATH]-1\lt \frac{3x+4}{x-6}[/MATH] and [MATH]\frac{3x+4}{x-6}\lt 1[/MATH]? If so, then do it, and show what you find. The first step is to add 1 to both sides of the first one and rewrite [MATH]\frac{3x+4}{x-6} + 1[/MATH] as a single fraction.

Jomo's case method can be used, but it takes a lot of work, and what you've shown isn't close.

 

[Steven G]

So would this be helpful as there could be a proof by contradiction for either conjecture to deduce whether x is actually positive or negative?

Say −1 < 3x + 4/ x − 6 < 1
suppose

So would this be helpful as there could be a proof by contradiction for either conjecture to deduce whether x is actually positive or negative?

Say −1 < 3x + 4/ x − 6 < 1
suppose (x-6) > 0 and suppose ( x − 6 )<0
The (x-6) > 0 cannot be true by contradiction --> x> 6
3x + 4< x − 6 , which is incorrect
so we know x-6 has to be -ve.

The (x-6) > 0 cannot be true by contradiction --> x> 6
3x + 4< x − 6 , which is incorrect
so we know x-6 has to be -ve.

suppose (x-6) > 0 and suppose ( x − 6 )<0. You can NOT assume both at the same time. SUPPOSE in case 1 you assume x-6>0. You do your legal math and come up with -2<x<8. Since you got this answer based on your assumption that x-6>0 (ie x>6) then the answer in this case will be 6<x<8. Do you see this?

Suppose in case 2 you assume that x-6<0, ie x<6 and you arrive at x>4. So there is no solution for this case. Do you see this?

The final answer for be 6<x<8. Is this clear.

Now sometimes doing it this way will save lots of times and other times it will not save you any time. After getting experience you will learn when to use it.

Try solving x/(x-2) > 4

 

[Steven G]

So would this be helpful as there could be a proof by contradiction for either conjecture to deduce whether x is actually positive or negative?

Say −1 < 3x + 4/ x − 6 < 1
suppose (x-6) > 0 and suppose ( x − 6 )<0
The (x-6) > 0 cannot be true by contradiction --> x> 6
3x + 4< x − 6 , which is incorrect
so we know x-6 has to be -ve.

I have no idea at all why you concluded that x-6 has to be negative. But I am sure that you can use that same exact knowledge to show that x-6 must be positive. As a result x=0 is the only possible solution left. Do you think that is always the case?