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.