h(x)=In abs (*x+1)/(3x-6)(x-4)

cool. nice function.

need derivative

d/dx ln(u) = 1/u du/dx

giving us

1/(x+1)

(3x-6)(x-4) = 3x^2 - 18x + 24

now we have f/g so (f/g)' = (f'g-fg')/g^2

= [1/(x+1) * (3x^2 - 18x + 24) - ln|x+1| * (6x^2 - 18)]/(3x^2 - 18x + 24)^2

I guess you can simplify that, but I don't see much wiggle room

The function you provided is h(x) = ln |(x+1)/(3x-6)(x-4)|.

To evaluate this function at a specific value of x, you can follow these steps:

1. Start by substituting the given value of x into the function h(x).

2. Simplify the expression inside the absolute value bars, (x+1)/(3x-6)(x-4), using the given value of x.

3. Evaluate the natural logarithm (ln) of the resulting expression.

Here's an example to illustrate the process:

Let's say you want to evaluate h(x) at x = 2.

1. Start by substituting x = 2 into h(x):

h(2) = ln |(2+1)/(3(2)-6)(2-4)|

2. Simplify the expression inside the absolute value bars:

h(2) = ln |(3)/(0)(-2)|

Notice that (3)/(0) is undefined. In this case, h(2) is also undefined.

It's important to note that the function h(x) might have different values and properties for different ranges of x.