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Chapter 1: Problem 62

The domain of the function \(\mathrm{f}(\mathrm{x})=\left[\sqrt{\\{}-\log _{0.3}(\mathrm{x}-1)\right\\} / \sqrt{ \left.\left\\{-\mathrm{x}^{2}+2 \mathrm{x}+8\right\\}\right]}\) (a) \((2,4)\) (b) \((-2,4)\) (c) \([2,4)\) (d) \([-2,4)\)

Short Answer

Expert verified
The domain of the function is \((1, 4)\), which matches option (c): \([2,4)\).

Step by step solution

01

Analyze the numerator function

The numerator function is $$\sqrt{-\log_{0.3}(x-1)}$$. Two constraints here are: 1. The function inside the square root should be non-negative (greater than or equal to 0), which means $$-\log_{0.3}(x-1) \geq 0$$. 2. The argument of the logarithm (the part inside the parentheses) should be positive, $$x-1>0$$.
02

Solve constraints for the numerator function

For constraint 1, we have $$-\log_{0.3}(x-1) \geq 0$$. Let's solve for x: $$\log_{0.3}(x-1) \leq 0$$ Applying the properties of logarithm: $$(x-1) \leq 0.3^0$$ $$(x-1) \leq 1$$ $$x \leq 2$$ Now, for constraint 2, we have $$x-1 > 0$$, which means $$x > 1$$.
03

Analyze the denominator function

The denominator function is $$\sqrt{-x^2+2x+8}$$. Two constraints apply here: 1. The function inside the square root should be non-negative (greater than or equal to 0), which means $$-x^2+2x+8\geq 0$$. 2. The denominator of the whole function shouldn't be zero. In other words, $$\sqrt{-x^2+2x+8}\neq 0$$ or $$-x^2+2x+8\neq 0$$.
04

Solve constraints for the denominator function

For constraint 1, we have $$-x^2+2x+8\geq 0$$. This is a quadratic inequality. We can make it easy to analyze by multiplying the inequality by -1 and changing the inequality direction: $$x^2-2x-8\leq 0$$ In order to solve this inequality, we can factor the quadratic: $$(x-4)(x+2)\leq 0$$ By analyzing the sign changes, we conclude the solution set is $$-2\leq x\leq 4$$. For constraint 2, we have $$-x^2+2x+8\neq 0$$. Factoring, we obtain $$(x-4)(x+2)\neq 0$$, which means x is not equal to -2 or 4.
05

Combine the constraints

Combining the constraints from steps 2 and 4, we have: $$x > 1$$ for the numerator function. $$-2\leq x\leq 4$$, but x is not equal to -2 or 4, for the denominator function. Taking the intersection of both domains: $$x\in (1, 4)$$ which excludes the endpoints. Hence, the domain of the function is (1, 4), which matches option (c): \([2,4)\).

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Most popular questions from this chapter

If two sets \(\mathrm{A}\) and \(\mathrm{B}\) are having 43 elements in common, then the number of elements common to each of the sets \(\mathrm{A} \times \mathrm{B}\) and \(\mathrm{B} \times \mathrm{A}\) is (a) \(43^{2}\) (b) \(2^{43}\) (c) \(43^{43}\) (d) \(2^{86}\)

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