To find the domain of this function why don't we consider the numerator?
In the function f(x) = (4-x)/2+log base 2 of x
to find the domain we set the denominator different from zero and x>0 cause of the log but why dont we set 4-x different from zero? please help me clarify this doubt, thanks in advance
1 Answer
Let $f(x)=\frac{4-x}{2}+\log_2 ^{x}$.
Since $\log_2 ^{x}$ is not defined for $x\leq 0$, we must have $x>0$. On the other hand $\frac{4-x}{2}$ is defined for all $x$ (note that there is no division by zero). So the domain of $f$ is $x>0$ or
\[(0,\infty).\]
You do not set $4-x$ to be zero, because $4-x$ in not in the denominator. It is fine to devide 0 by any number, but we should avoid dividing by zero. In the expression $\frac{4-x}{2}$ you never divide by zero.

405
Join Matchmaticians Affiliate Marketing
Program to earn up to 50% commission on every question your affiliated users ask or answer.
- 1 Answer
- 145 views
- Pro Bono
Related Questions
- Minimizing the cost of building a box
- piceiwise function, graph
- Volume of the solid of revolution
- Get the volume and surface area of the paraboloid $z=4-x^2-y^2$ cut by the plane $z=4-2x$
- Functions + mean value theorem
- Evaluate $\int_C (2x^3-y^3)dx+(x^3+y^3)dy$, where $C$ is the unit circle.
- Evaluate $\int \sin x \sqrt{1+\cos x} dx$
- Is $\int_1^{\infty}\frac{x+\sqrt{x}+\sin x}{x^2-x+1}dx$ convergent?
Domain means where (= for which x) the function is defined (= where it can be computed). The fraction (4-x)/2 can be computed for any x, no problem when it's negative or zero. But the second tem uses the log function, which is only defined for positive arguments, but not when x <= 0. (Check the graph of log(x) - you can ask google "plot log(x)" : it goes to minus infinity when x approaches 0 from the right.) So we need x > 0. (The sum of the two terms is defined whenever both are defined.)