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Find the domain of the real-valued function \[f(x)=\sqrt{-6x^2+11x-4}.\] Give the endpoints in your answer as common fractions (not mixed numbers or decimals).

Guest Mar 15, 2018
 #1
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\(f(x)=\sqrt{-6x^2+11x-4}\)

 

The domain is all real values of  x  such that

 

-6x2 + 11x - 4  ≥  0

 

To solve this inequality, let's first solve this equation...

 

-6x2 + 11x - 4  =  0       Using the quadratic formula...

 

x  =  \({-11 \pm \sqrt{11^2-4(-6)(-4)} \over 2(-6)}\,=\,{-11 \pm 5 \over -12}\)

 

x  =  \(\frac12\)         or        x  =  \(\frac{4}{3}\)

 

Since the x values that make  -6x2 + 11x - 4  equal  0  are  \(\frac12\)   and   \(\frac{4}{3}\) ,

(and since  -6x2 + 11x - 4  is continuous....)

the x values that make  -6x2 + 11x - 4  greater than or equal to  0  will be either be...

those between \(\frac12\)  and  \(\frac{4}{3}\)  ,  that is,  those in the interval  \([\frac12,\frac43]\)

OR

those outside of that interval, that is, those in the interval  \((-\infty,\frac12]\cup[\frac43,\infty)\)

 

To determine which...

 

Let's test an  x  value in the interval  \([\frac12,\frac43]\) .  Does  x = 1  make  -6x2 + 11x - 4  ≥  0    ?

-6(1)2 + 11(1) - 4  ≥  0

1 ≥ 0      true

 

Let's test an  x  value in the interval  \((-\infty,\frac12]\cup[\frac43,\infty)\) . Does  x = 0  make  -6x2 + 11x - 4  ≥  0 ?

-6(0)2 + 11(0) - 4  ≥  0

-4 ≥ 0     false

 

So the  x  values that make  -6x2 + 11x - 4  ≥  0  are those in the interval  \([\frac12,\frac43]\)

The domain of  f(x)  is  \([\frac12,\frac43]\)

hectictar  Mar 15, 2018

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