Is #1 / ( x )^2# strictly greater than # 1/(x)# for all x?

2 Answers
Oct 22, 2015

Answer:

No

Explanation:

#1/x^2 <= 1/x if x>=1#
#color(white)("XXX")#for example #1/(2^2) = 1/4 < 1/2#

Oct 22, 2015

Answer:

See explanation.

Explanation:

If #x<0#:

When #x# is negative, #1/x# will be negative, but #1/x^2# will always be positive (because if #x inRR#, then #x^2>=0#). Therfore, #1/x^2# will be greater than #1/x#.

If #x=0#:

Both expressions will be undefined. Therefore, you can't compare them.

If #0"<"x<1#:

When #x# is between 0 and 1, #1/x^2# will be greater than #1/x#.

If #x=1#:

Both expressions will be equal (they will both be 1).

If #x>1#:

When #x# is greater than 1, #1/x# will be greater than #1/x^2#. This is because in #1/x#, you are dividing 1 by #x# only once, while in #1/x^2#, you are dividing 1 by #x# twice.