Swapnil Saxena
Last Activity: 13 Years ago
Since root is defined only for >=0 and 1/x is defined for not= 0 thus 4x-|x^2 - 10x +9|> 0
However 4x-|x2 - 10x +9| can take two forms
Case I : For (x2 - 10x +9) >0 ==> (x-1)(x-9)>0 , it is equal to 4x-(x2 - 10x +9) => -x2+14x-9
Case II : For (x2 - 10x +9) <0 ==> (x-1)(x-9)<0 it is equal to 4x+(x2 - 10x +9) ==> x2-6x+9
Evaluating case I :
Now we ve to find when (x-1)(x-9)>0 ,so that the equation takes form -x2+14x-9
For this, either (x>1 and x>9 ==> x>9) or (x< 1 and x< 9 ==> x<1_
So for (-infinity, 1) and (9, infinity) , ths expression can take form -x2+14x-9,
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In such a situation, finding when -x2+14x-9 > 0 ,
Using Wavy curvy method, x is (7-2root(10),7+2root(10))
So using intersection of the two domains , between (7-2root(10),1)U(9,7+2root(10)),
4x-|x^2 - 10x +9|> 0
Evaluating case II:
Now we ve to find when (x-1)(x-9)<0 ,
for this purpose (x<1) and (x>9) , not possible
for this purpose (x>1) and (x<9) ,thus for (1,9), the equation will take form x2-6x+9
Now we have to evaluate when x2-6x+9 >0,
Again using wavy curvy, x is always positive for this equation
So x can belong to (1,9)
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After Evaluating both the case we find that domian can be (7-2root(10),1) U (1,9) U (9,7+2root(10)) =
( 7-2root(10) , 7+2root(10))