1.FInd the Range of y=log₃(5+4x-x²)

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