Question

Determine domain and codomain for f(x) = ln(x^2 - 6x)​

Answer 1

The domain:

Interval:
`(-\infty,0)\cup(6,\infty)`

Inequality: x<0 or x>6

Words: x is less than 0 or greater than 6.

The range:

Interval:
`(-\infty,\infty)`

Inequality:
`-\infty<y<\infty`

Words: All real numbers.

Explanation:

For the natural log (Ln) to exist, the inside must be positive.

So the domain can be found by solving the following:

`x^2-6x \text{ is positive }`

`x^2-6x>0`

Let's factor:

`x(x-6)>0`

`y=x^2-6x` is a faced up parabola with x-intercepts x=0 and x=6. This means it is positive when x<0 or when x>6. Those are the parts with the curve of the parabola is above the x-axis.

So the domain is x<0 or x>6.

Interval notation if you prefer would be:
`(-\infty,0)\cup(6,\infty)`.

`y=\ln(x^2-6x)` as equivalent exponential form
`e^(y)=x^2-6x`.

So let's look at the parabola one more time....

`y=x^2-6x` has it's minimum occur halfway between the x-intercepts we found earlier. Parabolas are symmetric about their axis of symmetry which the vertex lays on. So the halfway point of x=0 and x=6 is x=3. The vertex occurs at x=3.

To find the corresponding y-coordinate we can replace x with 3:

`y=3^2-6(3)=9-18=-9`.

The lowest point is -9 since the parabola is opened up.

So we know that
`e^y\ge -9`.

We also know every exponential function is greater than 0 so the intersections of what I just mentioned about our
`e^y` and all exponential functions is just >0.

`e^y>0` for all
`y`

So the range is all real numbers.

As an inequality:
`-\infty<y<\infty`

As an interval:
`(-\infty,\infty)`.