Question

Find out the values of x which satisfy
x^3(x-1)(x-2)^2>0
the answer given is (-infinity,0) U (1,2) U ( 2, infinity) .. please show the steps to solve the problem.

Answer 1

=taking values of x where function becomes 0 we can find the lowest limit actually :)
the numbers axcept 1 and 2 are all satisfying the expression either positive or negitive

Answer 2

hey x satisfies 2 in this expression first expand and then substitute.....

Answer 3

no x does not satisfy 2

Answer 4

if u put x=2 all expression will be 0 and 0>0 is wrong so listion dont putt x=1 or x=2 ..... other real numbers can be ur answers u can check ur self or if u have any number in dobt i can make it clear for u

Answer 5

X cannot be 0, 1 or 2, so anything else satisfies the equation. so x is the set of reals, but not equal to 0, 1 or 2.

Answer 6

My favourite way of solving this sort of thing is to take the factors of the expression one by one...
But first, I need you to identify what are called the 'critical numbers' of this expression.

Critical numbers would be the values of x which would make the left side of your inequality equal to zero...

Answer 7

x=0, x=1, x=2 wud make the left side 0
then...

Answer 8

this isn't your question? @rsadhvika
Anyway, we make a table :)
\[\Large \left[\begin{matrix}& -\infty & & 0 & & 1 & & 2 & & +\infty\\x^3&&&&&&&&\\(x-1)&&&&&&&& \\(x-2)^2 \\x^3 (x-1)(x-2^2)\end{matrix}\right]\]

Answer 9

yeah i have another method, but since its your fav method, want to see it... .and since the asker is not here, dont want to miss the method :o

Answer 10

this is for beginners, though... :)
anyway
The critical number associated with \(\large x^3\)
\[\Large \left[\begin{matrix}& -\infty & & 0 & & 1 & & 2 & & +\infty\\x^3&&&0&&&&&\\(x-1)&&&|&&&&& \\(x-2)^2&&&|&&&&& \\x^3 (x-1)(x-2^2)\end{matrix}\right]\]

Answer 11

@rsadhvika you can show your method as well

Answer 12

is zero, sorry.

Answer 13

after this ima post :)

Answer 14

Maybe your method is easier, @rsadhvika ?
I wouldn't want to complicate things for @digitalmonk

Answer 15

Anyway, before 0, \(\large x^3 \) takes negative values, and positive values after 0
\[\Large \left[\begin{matrix}& -\infty & & 0 & & 1 & & 2 & & +\infty\\x^3&-&-&0&+&+&+&+&+&+\\(x-1)&&&|&&&&& \\(x-2)^2&&&|&&&&& \\x^3 (x-1)(x-2^2)\end{matrix}\right]\]

Answer 16

oh each factor we will check and apply -x- = + ... i see looks pretty straight forward and cool :)

Answer 17

The critical number associated with (x-1)
[the x-value that would make x - 1 = 0]
is 1
\[\Large \left[\begin{matrix}& -\infty & & 0 & & 1 & & 2 & & +\infty\\x^3&-&-&0&+&+&+&+&+&+\\(x-1)&&&|&&0&&& \\(x-2)^2&&&|&&|&&& \\x^3 (x-1)(x-2^2)\end{matrix}\right]\]
Before 1, (x-1) takes negative values, after 1, x-1 takes positive values...
\[\Large \left[\begin{matrix}& -\infty & & 0 & & 1 & & 2 & & +\infty\\x^3&-&-&0&+&+&+&+&+&+\\(x-1)&-&-&-&-&0&+&+&+&+ \\(x-2)^2&&&|&&|&&& \\x^3 (x-1)(x-2^2)\end{matrix}\right]\]

Answer 18

Finally, for \(\large(x-2)^2\)
The critical number fot it is 2, but since it's squared, everywhere else, it's going to be positive...
\[\Large \left[\begin{matrix}& -\infty & & 0 & & 1 & & 2 & & +\infty\\x^3&-&-&0&+&+&+&+&+&+\\(x-1)&-&-&-&-&0&+&+&+&+ \\(x-2)^2&+&+&+&+&+&+&0&+&+ \\x^3 (x-1)(x-2^2)\end{matrix}\right]\]

Answer 19

Product of the three signs go at the bottom, just remember that - times - is +
\[\Large \left[\begin{matrix}& -\infty & & 0 & & 1 & & 2 & & +\infty\\x^3&-&-&0&+&+&+&+&+&+\\(x-1)&-&-&-&-&0&+&+&+&+ \\(x-2)^2&+&+&+&+&+&+&0&+&+ \\x^3 (x-1)(x-2)^2&+&+&0&-&0&+&0&+&+\end{matrix}\right]\]

And now just look for the intervals where the prouct is positive, since that's what you needed...

Answer 20

ok got you @terenzreignz ..thanks a lot..... @rsadhvika please show me your method as well when you get time.

Answer 21

i wud try something like this :-
x^3(x-1)(x-2)^2 > 0

Observation 1 :
see that its 6 degree polynomial inequality

Answer 22

for even degree polynomials extremes goes in same direction, since leading coefficient is +ve, the extremes goes UP

and since it has 3 distinct zeroes, it cuts/touches x axis at 3 places

so the domain where it stays UP wud be like this :-
(-inf, ) U (, inf)

Answer 23

next visualize its graph :- extremes going UP, adn touching/cuttings x-axis at 3 zeroes : 0, 1, 2