Domain of logarithmic function, find domain using interval notation

Question

I(x)= In (2 x² - 11 x+5)

Answer 1

My first instinct when I see a more involved problem like this is to break it into several simpler problems - after all, solving the parts is usually more straightforward than solving the whole. Before our breakdown, let's size up the problem: what is it asking? We're asking to find the domain of a function: in other words, the set of all possible inputs. The function we're dealing with here is a logarithmic one: the inverse of an exponential function.

When we talk about the inverse of a function, we can imagine taking the domain (input) and range (output) of a function and swapping them. For the exponential function f(x) = e^x, our domain is (-∞, ∞), the set of all real numbers, and our range is (0, ∞) . In the inverse function f(x) = ln(x), by contrast, our range is (-∞, ∞) and our domain is (0, ∞) . This puts the restriction on our domain that x > 0.

In the case of this problem, the input x has been replaced with the input 2x²-11x+5 , so our condition for the domain is now that 2x²-11x+5 > 0. To find the exact conditions on x here, it might help to factor our polynomial expression first. We can do this pretty handily by breaking apart the middle term and grouping:

2x²-11x+5

2x²-10x-x+5

2x(x-5)-(x-5)

(2x-1)(x-5)

The inequality for our domain condition now becomes (2x-1)(x-5) > 0. If the product of two numbers is greater than zero, we have a few possibilites: either both numbers are positive, or both numbers are negative. In math speak, this means that either (1) 2x-1 > 0 and x-5 > 0 OR (2) 2x-1 < 0 and x - 5 < 0. Let's solve each of these inequalities for x and try to make sense of the results.

(1) 2x - 1 > 0 and x - 5 > 0 → x < 1/2 and x < 5

(2) 2x - 1 < 0 and x - 5 < 0 → x > 1/2 and x > 5

These results split the number line into 3 intervals: (-∞, 1/2), (1/2, 5), and (5, ∞). To see which ones of these fulfill our domain condition, we can pick a test value from each and make sure the resulting inequality is true:

(-∞, 1/2): Picking the value 0 makes our inequality (2·0-1)(0-5) > 0 → (-1)(-5) > 0 → 5 > 0. True

(1/2, 5): Picking the value 1: (2·1-1)(1-5) > 0 → (1)(-4) > 0 → -4 > 0. False

(5, ∞): Picking the value 6: (2⋅6-1)(6-5) > 0 → (11)(1) > 0 → 11 > 0. True

Our domain must only be defined in the intervals that fulfill the condition (2x-1)(x-5) > 0, which are (-∞, 1/2) and (5, ∞).

Answer: (-∞, 1/2) ∪ (5, ∞)

Answer 2

l(x) = ln(2^{x2 - x + 5}) Domain is (-oo, oo)