1. remember that the denominator √(36 - x^2) cannot be zero or negative,
so shouldn't -6 < x < +6 ??
2. log6 [(x-11)(x-6)] = 2
(x-11)(x-6) = 6^2
x^2 - 17x + 66 = 36
x^2 - 17x + 30 = 0
(x-15)(x-2) = 0
x = 15 or x = 2
A lot of students would stop here and think they have the right answer, but remember that we can only take logs of positive numbers, so looking at our original we can see clearly that x > 11
so x = 15 is the only answer.
1.) Find the domain and range for:
f(x)= 4x-20/ the square root of 36-x^2
2.) Solve the equation:
log6 (x-11) + log6 (x-6) = 2
3.) Solve the inequality:
x^2 - 2x - 3 is greater than 0
2 answers
3. x^2 - 2x - 3 > 0
(x-3)(x+1) > 0
"critical" values are x = 3 and x = -1
try a number < -1, say x = -5
statement: (-8)(-4) > 0 ? YES
try a number between -1 and 3, say x = 0
statement: (-3)(1) > 0 False!
try a number > 3, say x = 5
statement: (2)(6)>0 YES
so x < -1 OR x > 3, x any real number.
(x-3)(x+1) > 0
"critical" values are x = 3 and x = -1
try a number < -1, say x = -5
statement: (-8)(-4) > 0 ? YES
try a number between -1 and 3, say x = 0
statement: (-3)(1) > 0 False!
try a number > 3, say x = 5
statement: (2)(6)>0 YES
so x < -1 OR x > 3, x any real number.
2 answers