Problem

$73-78=$ Domain Find the domain of the function. …

05:44

Problem 76 Hard Difficulty

$73-78=$ Domain Find the domain of the function.
$$g(x)=\ln \left(x-x^{2}\right)$$

Answer

$D_{g}=(0,1)$

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Algebra

Algebra and Trigonometry

Chapter 4

Exponential and Logarithmic Functions

Section 3

Logarithmic Functions

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Video Transcript

I want to note with this problem is that the natural log is equal. Thio do log rhythm with the of e, which is a constant. So these two expressions here are equal to one another. The next thing that we want to note is that the natural log of a component is tin is just the inverse function of E to that same expression. So what we want to do is find the domain of this function. So we want to find the values where this part here X minus X squared is greater than zero. And the reason we want to do that is because we, uh, know that e take into a negative value. So we'll start with negative. One is equal to one over each that value so each the negative first power is one over e. And even as you take me to even lower numbers so you can take each the negative 100 for example, this will be equal to one over you to the 100 and you can keep going and this number here we'll get smaller and smaller and smaller, but it will always be greater than zero. Uh, it can be very, very small, but it will always remain positive who will always be great in the zero. So we know that this expression up here needs to be greater than zero. So using that knowledge, we can to find the domain of the function as X minus X squared must be greater than zero. But we can't just stop here because this isn't a fully simplified expression. So the first thing that we're gonna want to do is I would recommend factoring because we can see that both of these expressions here contain an ex. So using the distributive property, we can take the X out and have X times one minus X must be greater than zero. And now we know that both of these components X and what X must both be greater than zero so that when their multiplied together, their solution is greater than zero. So we can separate that out and have X is greater than zero here, and one minus X is greater than zero. And this expression over here is already simplified. But to solve for X on this expression, we can add X to both sides. So an extra both sides and get one is greater than zero plus x which is just equal to so here we have two bounds for X. We know that X must be greater than zero, but that one must be greater than X. So we can write out are solution up here as the domain of the function Ah must need to find where zero is less than X and it is Let that one whoops. So essentially what we're working with freshen that states that X must be bounded between zero and one. And now that we found that we can, uh, check on a graphing calculator to see if just as a double check. And I did this earlier on Dismas. So we will insert this photo here and we can see the graft. And as we can see in the graph, we can note that this red line here is bounded between X equals zero and X equals one. It does not cross either one of these vertical lines here, so we know that our solution is correct.

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