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1.FInd the Range of y=log 3 (5+4x-x 2 )

1.FInd the Range of y=log3(5+4x-x2)

Grade:11

2 Answers

sudhir pal
askIITians Faculty 26 Points
8 years ago
solved the problem analytically. since the function inside the log is parabola opening downward it will take max value at its vertex which is (2,9). other details are in the picture I'm attaching.
Thanks & Regards
Sudhir,
askIITians Faculty
Qualification.
IIT Delhi
153-1635_Photo1730.jpg
Rishi Sharma
askIITians Faculty 646 Points
2 years ago
Dear Student,
Please find below the solution to your problem.

y is defined for -x^2+4x+5>0 or
+x^2-4x-5 <0 ,factorise to get
(x-5)(x+1)<0 this is true in the interval
A= { x / , -1 < x <5 }
So y is defined in that domain.
at x = -1 and x= 5, are asymtopes .
As x tends to -1 from RHS y tends to - inf
As x tends to +5 from LHS y tends to - inf
So the lower bound of the range is - inf.
Within that domain there must be a maximum ,and possible roots.
First find roots. These occur when y=0 or 5+4x-x^2 = 1
x^2-4x-4 =0
x= 2-2√2 = -0.828 . or x 2+2√2 =4.828
To find max differentiate y= ln(5+4x-x^2)
Differentiate : y' = (-2x +4)/(5+4x-x^2) =0 , x= 2 at max/ min.
2nd derivative <0 implies max ,otherwise min.
y'' = [(5+4x-x^2)(-2)-(4-2x)(4-2x)]/( v^2)
At x =2 , y" <0 and y = ln 9 ,max at (x,y) =( 2, ln 9 ).
So y starts at -inf ,at x>-1 (from RHS) increases until it finds a root at x = -0.824 .
Continues unti!l it reaches a max . at ln(9) when x =2 .
Starts to decrease to find a root at x = 2+√2 = 4.824 and then continues on it's merry way to -inf. as it tends to x=5 from the LHS

Range =[-inf,ln 9]

Thanks and Regards

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