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# Help please

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

Jul 19, 2019

### 2+0 Answers

#1
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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
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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