How can an inequality be manipulated to show a specific range of values?

  • Thread starter Thread starter shawli
  • Start date Start date
  • Tags Tags
    Inequality
AI Thread Summary
The discussion focuses on how to manipulate the inequality -1 ≤ x ≤ 1 to demonstrate that 0 ≤ x² ≤ 1. The key point is that squaring any value of x within the range of -1 to 1 results in a non-negative value, hence the lower bound becomes 0. Additionally, since the maximum value of x in this range is 1, squaring it confirms that the upper bound remains 1. Visualizing the graph of x² over the interval clarifies this transformation. Understanding these properties of squaring helps in grasping the relationship between the two inequalities.
shawli
Messages
77
Reaction score
0
In an example in my textbook, it says the following:

"If -1 ≤ x ≤ 1, then 0 ≤ x2 ≤ 1. "


Can someone explain to me how to move from the first statement to the second statement please? I'm not quite sure how the -1 turned into a 0...
 
Physics news on Phys.org
Did you draw a graph of x^2 on the interval -1 ≤ x ≤ 1?
 
Ah right... it's staring me right in the face! So clear that I missed it haha. Thank you!
 

Thread 'Finding the number of ways to arrange identical balls in a circle (3 different colors)'

I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
View full post »

Thread 'Making the denominator of a certain fraction real'

Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
View full post »

Similar threads

Need help in manipulating rational absolute value inequalities
Replies
11
Views
2K
Solving for ##x## for a given inequality
Replies
3
Views
1K
Can this system of inequalities be solved for x?
Replies
5
Views
1K
Find the values of a real number ##a## for an inequality to hold
Replies
7
Views
2K
Solving an absolute value quadratic inequality
Replies
9
Views
2K
Solving an inequality involving absolute values
Replies
11
Views
2K
Doubt solving a polynomial inequality
Replies
4
Views
2K
Manipulating an inequality in the bisection method
Replies
2
Views
2K
Having problems in multiplying the inequalities
Replies
19
Views
2K
Find the range of the function ##f(x) = \frac{e^{2x}-e^x+1} {e^{2x}+e^x+1}##
Replies
8
Views
3K

Hot Threads

Recent Insights

Back
Top