To compare the functions \(f(x) = \sqrt{x+1} - 2\) and \(g(x) = \sqrt[3]{x+1} - 2\), we will analyze their domains, ranges, y-intercepts, and x-intercepts.
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Domain:
- For \(f(x) = \sqrt{x+1} - 2\), the expression under the square root must be non-negative. Therefore, \(x + 1 \geq 0\) or \(x \geq -1\). The domain of \(f\) is \([-1, \infty)\).
- For \(g(x) = \sqrt[3]{x+1} - 2\), the cube root function is defined for all real numbers, so the domain of \(g\) is \((- \infty, \infty)\).
- Conclusion: The domains are different.
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Range:
- For \(f(x) = \sqrt{x+1} - 2\), the minimum value occurs at \(x = -1\), giving \(f(-1) = \sqrt{0} - 2 = -2\). As \(x\) increases, \(f(x)\) increases without bound. Thus, the range of \(f\) is \([-2, \infty)\).
- For \(g(x) = \sqrt[3]{x+1} - 2\), as \(x\) approaches \(-\infty\), \(g(x)\) approaches \(-2\), and as \(x\) approaches \(\infty\), \(g(x)\) also approaches \(\infty\). Thus, the range of \(g\) is also \((-2, \infty)\).
- Conclusion: The ranges are not the same (the range of \(f\) includes \(-2\) while the range of \(g\) does not).
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Y-intercepts:
- For \(f(x)\), when \(x = 0\), \(f(0) = \sqrt{1} - 2 = -1\). So, the y-intercept of \(f\) is \((0, -1)\).
- For \(g(x)\), when \(x = 0\), \(g(0) = \sqrt[3]{1} - 2 = -1\). So, the y-intercept of \(g\) is also \((0, -1)\).
- Conclusion: The y-intercepts are the same.
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X-intercepts:
- For \(f(x)\), setting \(f(x) = 0\) gives \(\sqrt{x+1} - 2 = 0\) which leads to \(\sqrt{x+1} = 2\), squaring both sides results in \(x + 1 = 4\), thus \(x = 3\). So, the x-intercept of \(f\) is \((3, 0)\).
- For \(g(x)\), setting \(g(x) = 0\) gives \(\sqrt[3]{x+1} - 2 = 0\) which results in \(\sqrt[3]{x+1} = 2\). Cubing both sides gives \(x + 1 = 8\), thus \(x = 7\). So, the x-intercept of \(g\) is \((7, 0)\).
- Conclusion: The x-intercepts are different.
Final Answer:
The correct statement is: The graphs have the same y-intercept.
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