need clarification with inequalities (1 Viewer)

kangarulz

Procrastinaton
Joined
Mar 23, 2005
Messages
144
Gender
Female
HSC
2006
with the inequalities with results like: (x-2)(x+1)>0 or (x-1)(x+3)(x-2)<0, etc how do you when to reverse the sign? and also does it make a difference to the result if the question asks (2-x) instead of (x-2) (where the x and the number has switched positions)
 

B35tY

Member
Joined
Feb 1, 2005
Messages
262
Location
Sydney
Gender
Male
HSC
2006
I'm not too sure what you're asking, but i'll try to help.

To solve the inequality (x-2)(x+1)>0

Firstly, it should be apparent that the equation (x-2)(x+1)>0 has roots at x = 2 and x = -1

Then, by expanding the first terms of each bracket (getting x^2), you can see that the graph will be a normal parabola (as opposed to an upside down parabola, which would be the case if the coefficient of x^2 was negative).

Then, draw a quick graph and it's easy to see that the answer is x > 2 or x < -1

Of course, if you had (2 - x) instead of (x - 2) then the coefficient of x^2 would be -1 and then the answer would be -1<x<2 instead..

Did i help?
 

ThuanSUX

New Member
Joined
Aug 21, 2006
Messages
9
Gender
Male
HSC
2000
I agree with b35ty about drawing the parabola. From the factorized form you can mark your roots on the x-axis. From here draw it out (remember to watch out for the (2-x) case because the parabola is heading downwards).

If you want f(x) < 0, then it's the part of the parabola under the x-axis
If you want F(x) > 0, then it's the upper part.

There is a large difference between (x-2) and a (2-x) factor. Although the roots are the same, the curve is up-side-down. This means your answers to the inequality should have the opposite signs.

Eg: (x-2)(x+3) > 0
Ans: x<-3 and x>2

Eg: (2-x)(x+3) > 0
Ans: -3 < x < 2
 

Riviet

.
Joined
Oct 11, 2005
Messages
5,593
Gender
Undisclosed
HSC
N/A
ThuanSUX said:
If you want f(x) < 0, then it's the part of the parabola under the x-axis
If you want F(x) > 0, then it's the upper part.
This isn't true for all continuous functions.
Instead, you could think of it like this:
Say you have some function f(x)>0. Sketch the graph as stated by others with the x-intercepts and any important info that might be useful. Ask yourself, "for what values of x is the curve above the x-axis?" Determine these x-values and that will give your solution.

In general,

For f(x)>0, determine the x-values for which f(x) is strictly above the x-axis.

For f(x)>0, determine the x-values for which f(x) touches OR is above the x-axis.

For f(x)<0, determine the x-values for which f(x) is strictly below the x-axis.

for f(x)<0, determine the x-values for which f(x) touches OR is below the x-axis.
 

ThuanSUX

New Member
Joined
Aug 21, 2006
Messages
9
Gender
Male
HSC
2000
Riviet said:
In general,

For f(x)>0, determine the x-values for which f(x) is strictly above the x-axis.

For f(x)>0, determine the x-values for which f(x) touches OR is above the x-axis.

For f(x)<0, determine the x-values for which f(x) is strictly below the x-axis.

for f(x)<0, determine the x-values for which f(x) touches OR is below the x-axis.

That's what I meant, but I couldn't figure out how to do the greater/equal and lesser/equal symbol and cbf looking.

Riviet said:
This isn't true for all continuous functions.
I believe what I said was correct. For f(x) > 0, you simply use the > sign instead of >, and likewise for lesser than. :)
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top