Question

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

Answer 1

Answer:

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.

Step-by-step 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: $(-\infty,\infty)$.