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Chapter 2: Problem 63

Find the domain of the expression. Use a graphing utility to verify your result. $$\sqrt{x^{2}-9 x+20}$$

Short Answer

Expert verified
The domain of the expression \( \sqrt{x^{2}-9x+20} \) are x-values for which x is within the intervals (-鈭�, 4] and [5, 鈭�).

Step by step solution

01

Set the inequality

Start by setting the expression inside the square root greater than or equal to zero: \(x^{2}-9x+20 \geq 0\)
02

Factor the quadratic equation

Factor the quadratic equation to solve for x values: \((x - 4) * (x - 5) \geq 0\)
03

Find the critical points

The critical points occur where the expression is equal to zero: x = 4, x = 5. These points divide the real number line into three intervals: (-鈭�, 4), (4, 5), and (5, 鈭�).
04

Test values in each interval

Test values from each interval in the factored inequality. Choose x = 0 from (-鈭�, 4), x = 4.5 from (4, 5) and x = 6 from (5, 鈭�). The inequality is satisfied when x is in (-鈭�, 4], or x is in [5, 鈭�). Any x-value from these two intervals will make the expression inside the square root non-negative.
05

Verify with a graphing utility

Plot the function \(y = \sqrt{x^{2}-9x+20}\) using a graphing utility to verify. You should see that the graph exists wherever x is in the intervals (-鈭�, 4] or [5, 鈭�)

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Most popular questions from this chapter

Write the polynomial as the product of linear factors and list all the zeros of the function. $$g(x)=x^{4}-4 x^{3}+8 x^{2}-16 x+16$$

Write the polynomial as the product of linear factors and list all the zeros of the function. $$g(x)=x^{2}+10 x+17$$

Use Descartes's Rule of Signs to determine the possible numbers of positive and negative real zeros of the function. $$g(x)=2 x^{3}-3 x^{2}-3$$

Describe the error. $$\sqrt{-6} \sqrt{-6}=\sqrt{(-6)(-6)}=\sqrt{36}=6$$

Write the polynomial (a) as the product of factors that are irreducible over the rationals, (b) as the product of linear and quadratic factors that are irreducible over the reals, and (c) in completely factored form. $$f(x)=x^{4}+6 x^{2}-27$$
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