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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).

Mar 15, 2018

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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  $$[\frac12,\frac43]$$

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Mar 15, 2018