1. ## Domain and range

obtain the domain and range of ln(g(x))??

g(x)=6+4x-2x2

2. Hi

g(x)=-2x²+4x+6 = -2(x+1)(x-3)
It must be strictly positive because ln(a) is defined for a>0
The domain is therefore ]-1,3[

g(x)=-2x²+4x+6 is a parabola oriented towards negative y
Maximum is obtained for x=1 and g(1)=8
Range is therefore ]-oo,3ln(2)]

3. Originally Posted by running-gag
Hi

g(x)=-2x²+4x+6 = -2(x+1)(x-3)
It must be strictly positive because ln(a) is defined for a>0
The domain is therefore ]-1,3[

g(x)=-2x²+4x+6 is a parabola oriented towards negative y
Maximum is obtained for x=1 and g(1)=8
Range is therefore ]-oo,3ln(2)]
Can you explain in details How can you find the range? ,I don't understand it

4. OK
ln(g(x)) is defined on ]-1,3[

On ]-1,3[ g(x) is increasing between -1 and 1 and decreasing between 1 and 3
Therefore ln(g(x)) is increasing between -1 and 1 and decreasing between 1 and 3

g(-1)=0 => ln(g(x)) has -oo for limit in -1
g(1)=8 => ln(g(1)) = ln(8) = ln(2^3) = 3 ln(2)
g(3)=0 => ln(g(x)) has -oo for limit in 3

5. Originally Posted by running-gag
OK
ln(g(x)) is defined on ]-1,3[

On ]-1,3[ g(x) is increasing between -1 and 1 and decreasing between 1 and 3
Therefore ln(g(x)) is increasing between -1 and 1 and decreasing between 1 and 3

g(-1)=0 => ln(g(x)) has -oo for limit in -1
g(1)=8 => ln(g(1)) = ln(8) = ln(2^3) = 3 ln(2)
g(3)=0 => ln(g(x)) has -oo for limit in 3

Thank you very much for Explaining

6. when I study again I have question in my mind

why we don't say that the range (0,ln8]

7. Because when x is close to -1 g(x) is close to 0 and we know that ln has a limit -oo in 0. Therefore ln(g(x)) has -oo as limit when x is close to -1

8. thank you very much .. now range will be very clear for me