What is the domain of f in f(x)=ln(x^2+4x-5) ?

Expert Answers
lemjay eNotes educator| Certified Educator

`f(x)= ln(x^2+4x-5)`

To determine the domain, apply the property of logarithm which is its argument is always positive `( ln a, agt0)` .

So, set x^2+4x-5 greater than zero.


Then, factor left side.


And set each factor equal to zero and solve for x.

For the first factor,




For the second factor,

`x - 1 = 0`

`x -1 + 1=0+1`


Next, determine the intervals in which it satisfy the inequality equation, using the values of x as the boundary of each interval.

For the first interval `(-oo,-5)` use the test value x=-6 and substitute it to the inequality equation.




`7gt0` (True)

For the second interval `(-5,1)` , use the test value x=0.



`-5gt0` (False)

And for the third interval `(1,+oo)` , use the test value x=2.



`7gt0` (True)

Hence, the domain of the given function is `(-oo,-5)uu(1,+oo)` .

oldnick | Student

funtion log(t)  is defined for  `t>0`

in this case `t=x^2+4x-5`  thus is enough to find for wich values of x   `t>0`

First step is to findthe root of equation:

`Delta= 16-4(-5)=36>0` 

thus has two differete real solutions



`x_1=1`     `x_2 = -5`

Then we can write  `x^2+4x-5=(x-1)(x+5)`  

Now this product is greater than zero if    `x<-5`  and `x>1`

so for this value   `t>0` , that is  function is defined.

note that in the  closed segment  [-5;1] funtion isn't deifned. (Both the point -5 and 1 include bicouse for this value the argument t =0)