VeritasPrepKarishma wrote:
vjsharma25 wrote:
I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values.
What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change?
Ok, look at this expression inequality: (x+2)(x-1)(x-7) < 0
Can I say the left hand side expression will always be positive for values greater than 7? (x+2) will be positive, (x - 1) will be positive and (x-7) will also be positive... so in the rightmost regions i.e. x > 7, all three factors will be positive. The expression will be positive when x > 7, it will be negative when 1 < x < 7, positive when -2 , x < 1 and negative when x < -2. We need the region where the expression is less than 0 i.e. negative. So either 1 < x < 7 or x < -2.
Now let me add another factor: (x+8)(x+2)(x-1)(x-7)
Can I still say that the entire expression is positive in the rightmost region i.e. x>7 because each one of the four factors is positive? Yes.
So basically, your rightmost region is always positive. You go from there and assign + and - signs to the regions. Your starting point is the rightmost region.Note: Make sure that the factors are of the form (ax - b), not (b - ax)...
e.g. (x+2)(x-1)
(7 - x)<0
Convert this to: (x+2)(x-1)
(x-7)>0 (Multiply both sides by '-1')
Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '-'
Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0.
Responding to a pm:
Quote:
i have a doubt in the highlighted region. U have said that it will be always +ve but in a bunel post he has asked to substitute the extreme values and if the f(x) is -ve then the right most of the inequality will be -ve.
Please clarify me on this.
Different people use different methods of solving problems. Both of us are correct. But you cannot mix up the methods. When you follow one, you have to follow that through and through.
When I say that the rightmost region will always be positive, it is after I make appropriate changes. Right below the highlighted portion, notice the note given:
Note: Make sure that the factors are of the form (ax - b), not (b - ax)...
I convert all factors to (ax - b) form. Now the rightmost region is positive by default.
Bunuel prefers to keep the factors as it is and check for the rightmost region.
What you would like to follow is your personal choice.
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