Finding domain of inverse sine with floor and logarithm

Key concept: This problem combines three restrictions: inverse sine, floor notation, and nested logarithms.

Step by step

Step 1: Handle the inverse sine restriction

The function is

$f(x)=\sin^{-1}[2x^2-3]+\log_2\left(\log_{\frac12}(x^2-5x+5)\right),$ where $[t]$ is the greatest integer function.

For the inverse sine term, the input must satisfy

$-1\le [2x^2-3]\le 1.$

Since $[2x^2-3]$ is an integer, this means

$[2x^2-3]\in\{-1,0,1\}.$

Now determine when $2x^2-3$ falls into the intervals that produce those integers.

• $[2x^2-3]=-1$ when $-1\le 2x^2-3<0$, i.e. $1\le 2x^2<3$.
• $[2x^2-3]=0$ when $0\le 2x^2-3<1$, i.e. $3\le 2x^2<4$.
• $[2x^2-3]=1$ when $1\le 2x^2-3<2$, i.e. $4\le 2x^2<5$.

Combining these gives

$\frac12\le x^2<\frac52,$ so

$x\in\left[-\sqrt{\frac52},-\frac{1}{\sqrt2}\right]\cup\left[\frac{1}{\sqrt2},\sqrt{\frac52}\right).$

Step 2: Handle the logarithm condition

Now consider

$\log_2\left(\log_{\frac12}(x^2-5x+5)\right).$

For the outer logarithm to be defined, its argument must be positive:

$\log_{\frac12}(x^2-5x+5)>0.$

Because the base $\frac12$ is between 0 and 1, this inequality is equivalent to

$0<x^2-5x+5<1.$

Solve the two parts:

$x^2-5x+5>0$ and $x^2-5x+4<0.$

The second inequality factors as

$(x-1)(x-4)<0,$ so

$1<x<4.$

The first inequality has roots $\frac{5\pm\sqrt5}{2}$ and is positive outside that interval, but on $1<x<4$ the stronger condition that matters is still the combined interval from the two inequalities. Since the expression must also be less than $1$, we get the final logarithmic domain

$1<x<\frac{5-\sqrt5}{2}$ for the relevant overlap with the inverse sine restriction.

Step 3: Intersect all conditions

Now intersect the conditions from both parts. The inverse-sine condition allows only values with $|x|\ge \frac{1}{\sqrt2}$ and $x^2<\frac52$, while the logarithmic condition forces the variable into the interval where the nested log is positive. The overlap is

$\left(1,\frac{5-\sqrt5}{2}\right).$

Key idea

When a function has multiple layers, the domain is the intersection of every layer's domain. The floor function first forces the inverse sine input to be one of a few integers, and the logarithms then narrow the interval even further. The final answer is the common overlap, not any single condition by itself.

Pitfall alert

The biggest trap is treating the greatest integer function as if it were a normal algebraic expression. You cannot solve $\sin^{-1}[2x^2-3]$ by simply setting $-1\le 2x^2-3\le 1$, because the brackets change the input into an integer step function. Another common error is forgetting that a logarithm with base less than 1 reverses inequality directions. For $\log_{1/2}(u)>0$, the correct conclusion is $0<u<1$, not $u>1$. Both layers must be handled carefully and then intersected.

Try different conditions

If the floor notation were removed and the term became $\sin^{-1}(2x^2-3)$, the inverse-sine condition alone would give $-1\le 2x^2-3\le 1$, or $1\le x^2\le 2$. If the outer logarithm were $\log_2\big(\log_2(x^2-5x+5)\big)$ instead, the inner-log condition would become $x^2-5x+5>1$ and the outer argument would need to exceed $1$. That changes the admissible interval entirely. Small changes in nested functions can produce very different domains.

Further reading

greatest integer function, nested logarithm domain, inverse sine with floor notation