What is the domain of f in 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.
For the second interval `(-5,1)` , use the test value x=0.
And for the third interval `(1,+oo)` , use the test value x=2.
Hence, the domain of the given function is `(-oo,-5)uu(1,+oo)` .
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:
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)