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Chapter 1: Problem 54

Describe the interval(s) on which the function is continuous. $$ f(x)=x \sqrt{x+3} $$

Short Answer

Expert verified
The function \(f(x) = x\sqrt{x+3}\) is continuous on the interval \([-3, \infty)\).

Step by step solution

01

Understand the domain of square root function

The square root function, \(\sqrt{z}\), is only defined for \(z \geq 0\). So, for our function \(f(x)=x\sqrt{x+3}\), we need the term under the square root, \(x+3\), to be greater or equal to zero.
02

Set up and solve the inequality

We need \(x+3 \geq 0\), which simplifies to \(x \geq -3\). So, the function should be okay to the right of -3, including -3.
03

Confirm the function works at x = -3

We need to make sure \(f(x)\) is defined at \(x=-3\). Plug \(x=-3\) into \(f(x)\), we have \(f(-3) = -3*\sqrt{-3+3} = 0\), not a undefined number, so -3 should be included in the domain.
04

Determine the interval on which the function is continuous

Based on the analysis, the function \(f(x)\) is continuous for all \(x \geq -3\). Thus, the interval of continuity is \([-3, \infty)\).

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