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\(\text{Compute the domain of the function $f(x)=\frac{1}{\lfloor x^2-7x+13\rfloor}.$}\)

 Jul 19, 2019
 #1
avatar+474 
+2

Since b^2-4ac = -3 for \(x^2-7x+13\), that means \(x^2-7x+13\) has no zeroes.

So, the domain of \(f(x)=\frac{1}{x^2-7x+13}\) is R.

 Jul 19, 2019
 #2
avatar+8579 
+3

x  can be all real numbers except those that make  \(\lfloor x^2 - 7x + 13\rfloor = 0\)

 

...which means...

 

x  can be all real numbers except those that make  \(0\ \leq\ x^2-7x+13\ <\ 1\)

 

To find the  x  values that make the inequality true, it helps to look at a graph of  y = x2 - 7x + 13:

 

 

We can see that there are no  x  values which make  y  equal  0 , just like Davis found.

But we can also see that there are some  x  values which make  y  fall between  0  and  1

 

Let's find what  x  values make  y  equal to  1

 

1  =  x2 - 7x + 13

                                               Subtract  1  from both sides of the equation.

0  =  x2 - 7x + 12

                                               Factor the right side of the equation.

0   =  (x - 3)(x - 4)

                                               Set each factor equal to zero and solve for  x

x - 3  =  0     or     x - 4  =  0

 

  x  =  3        or       x  =  4

 

Now we can see that if  x  is between  3  and  4,  y  will be between  0  and  1

 

So the domain is all real numbers except those exclusively between  3  and  4

 

In interval notation, the domain is  \((-\infty,3)\cup(4,\infty)\)_

 Jul 20, 2019

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