Ordered Powers
If − 1 < x < 0 , then what is the correct ordering of x , x 2 , x 3 , and x 4 ?
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6 solutions
Yeah the same way!
same way .
But, in the original question " If -1< x < 0, (not -1< x < 1 as you did.), then what is the ordering of x, x^2, x^3, x^4 ?"
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− 1 < x < 0 may be split into two cases, − 1 < x and x < 0 . It will be hard to compare the alternating value from the outset, so I make x positive, which gives 1 > − x and − x > 0 , or simply 0 < − x < − 1
If you know the property of exponents, however, it will be fairly safe to say that, for − 1 < x < 0 , then − 1 < x 2 n + 1 < x 2 m + 1 < 0 for positive integer n < m ( m → ∞ ) Similarly, with the same properties 0 < x 2 m < x 2 n < 1
All you have to do is replace x with a number
I did it that way letting x=-1/2.
I split up the odd exponents and even exponents.
The odd exponents will stay negative. The even exponents will be a positive value. So the odd exponents will come first in the inequality, then the even.
When you take any fraction between (-1,0), you can notice that multiplying that number by itself an odd number of times, you'll arrive at a value greater than that fraction.
Therefore x^3 is greater than x.
When you multiply a fraction between (-1,0), an even number of times, you will arrive at a positive value less than the original fraction. The more times you multiply this fraction, the smaller the product will be.
Therefore, x^4 is less than x^2.
To finalize,
x < x^3 < x^4 < x^2
Good solution.
Why is this level 2? Too simple
Lets replace x with − n 1 , x ≤ 0
When x squared,then ( − n 1 ) 2 = n 2 1
When x cubed,then ( − n 1 ) 3 = − n 3 1
And finally ( − n 1 ) 4 = n 4 1
Because these are fractions,so − n 1 < − n 3 1 < n 4 1 < n 2 1 or x < x 3 < x 4 < x 2
Details
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( − x ) n is negative if n is odd because ( − x ) 2 n + 1 = ( − x ) 2 n × ( − x ) = ( ( ( − x ) 2 ) n × ( − x ) = ( − x × − x ) n × ( − x ) = ( − 1 × x × − 1 × x ) n × ( − x ) = ( ( − 1 ) 2 × x 2 ) n × ( − x ) = ( x 2 ) n = x 2 n × − 1 × x = − x 2 n + 1 and positive if n is even because ( − x ) 2 n = ( ( ( − x ) 2 ) n = ( − x × − x ) n = ( − 1 × x × − 1 × x ) n = ( ( − 1 ) 2 × x 2 ) n = ( x 2 ) n = x 2 n
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Fractions is smaller when their denominator are bigger
I simply used − 2 1 as x, and then did x 2 , x 3 , x 4
Let 0 < − x < 1 (just rewrite it), cubing it gives 0 < − ( x 3 ) < − x < 1 (since any number between 0 and 1, when exponent, will has the decrease value) or simply − 1 < x < x 3 < 0
In similar manner, when squaring it, 0 < − x < 1 , it will be 0 < x 2 < 1 (another square for x 4 remains the same), but given the even-number exponents will give positive value, this will gives 0 < x 4 < x 2 < 1 Combining the two inequality gives − 1 < x < x 3 < 0 < x 4 < x 2 < 1