Question

I don't understand why I got this question wrong...?

Answer 1

The question was "Find the solutions to 0 ≥ x^4 – 1 by graphing y ≥ x^4 – 1 by hand."
And I put:
2. 0 >= ( x^4 – 1^4)
0 >= ( x^2 - 1^2)^2
0 >= [(x-1)(x+1)]^2
0 >= x ± 1
-x >= ± 1 / (-1)
± 1 >= x

My teacher graded my homework and said it was wrong

Answer 2

it is wrong

Answer 3

yes actually the correction is here.


You said (x^4-1^4) = (x^2-1^2)^2.

But (x^2-1^2)^2 = x4 + 1 - 2x^2. [(a-b)^2 = a^2 - 2ab + b^2, and not a^2-b^2)

Infact (x^4-1^4) = (x^2+1)(x^2-1)

=> (x^2+1)(x+1)(x-1) >= 0

so x^2 = -1, x = 1, x = -1.

but x^2 = -1 is an imaginary solution so actually we don't really take it.

Answer 4

well i dont understand why u solve it anallytacally
the problem says that
"Find the solutions to 0 ≥ x^4 – 1 by graphing y ≥ x^4 – 1 by hand
u just need to make the graph

Answer 5

this is a tricky question.

first, as lucky just posted x^4-1^4 is not the same as (x^2 - 1^2)^2

you do have a difference of squares a^2 - b^2 = (a-b)(a+b)

if you use that idea you eventually find
(x^2+1)(x+1)(x-1) ≤ 0

this will be less than 0 if it is negative.
x^2+1 will always be negative. so you want is
either
(x+1)<1 and (x-1) > 0 (their product will be negative) or
(x+1) > 1 and x-1 < 1

Answer 6

@phi actually x = root of -1..we know the value of iota? :P

Answer 7

from the u can see that the solution interval is -1<x<1

Answer 8

Thank you all so much!

Answer 9

**x^2+1 will always be positive