Is x^-2 < x^-1 for any negative value(s) of x? Why or why not?

Question

No stupid answers... college level algebra.

Answer 1

X^-2 compared to (-x)^-2 are both the same
X^-1 compared to (-X)^-1 are not
X^-2 is

1/X^2

1/X & -1/X

for all values X<0
the inequality becomes
1/X^2 < 1/X
and
assume X to be -1
1/(-1)^2 = 1
1/-1 = -1

1< -1 is false so the inequality is not valid.

Answer 2

Is x^-2 < x^-1 for any negative value(s) of x? Why or why not?

= 1/x^2 < 1/x
The first term is always positive for all x, however
the second term is positive when x is positive and negative when x is negative.

Therefore the inequality does not hold for x < 0.
A positive number cannot be less than a negative number.

Answer 3

x^(-2) < x^(-1)

First, convert these into having positive exponents.

1/x^2 < 1/x

Multiply both sides by x. Note that x is a negative value (given by the question), so this will switch the sign.

x/x^2 > 1

Now, we can cancel the left hand side.

1/x > 1

Multiply both sides by x; this will switch the sign (for multiplying by the negative number)

1 < x, which means x > 1. This contradicts the fact that x is a negative value, so this is never true.

Answer 4

No

If x< 0
then x^(-1) < 0
However x² > 0 so x^(-2) > 0

Therefore x^(-2) > x^(-1) for all x < 0

Answer 5

graph it and see if the negative values of
x^-2 are less than x^-1

Answer 6

x^-2 = 1/ (x^2) > 0
Where as
x^-1 = 1 / x < 0 (for x<0)

So the statement is false

Answer 7

(1/x)^2 < (1/x)

no

because

(1/-5)^2= a postive number

(1/-5) = a negative number

so if x is negative

x^-2 > x^-1

Answer 8

let x=-3
(-3)^-2=1/(-3)^2=1/9
x^-1=1/x=1/-3
x^-2>x^-1