Domain of a function with $\ln$

Question

Hy everyone, Im stuck on this assigment now :/

so the function is $f(x) = \ln(x - x^2 +12)$ and I need to get the domain of that function.

I know that $x - x^2 + 12 > 0$, because of $\ln$. then I seperated $x$'s on left and number on right

$x - x^2 > -12$ and I know that I can also write it as $x(1-x) > -12$

How do I get domain out of this :/ ?

Answer 1

So your problem is solving a second degree inequality.

Write it with a positive main coefficient (it is not a "must", but it is quite a "should"): $$x^2-x-12<0$$ Note that the RHS is $0$. Find the roots and factor the polynomial: $$(x+3)(x-4)<0$$ Now this product is negative only when the two factors have different signs. This is when $x$ is in the interval $(-3,4)$.

Answer 2

You can study the parabola, find its roots. There will be a small part where it is positive $\Delta=1+48=49=7^2$ Hence you have the roots and yu know it will be positive between them.

Answer 3

solving the inequality $$-x^2+x+12>0$$ we get $$-3<x<4$$ and for $y$ we obtain $$-\infty<y\le \log\left(\frac{49}{4}\right)$$