square-root-graph se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.
10 questions tagged with square-root-graph.
What is the real domain of (f(x)=\sqrt{16-x-2})?
A. ([-4,4])
वास्तविक वर्गमूल के लिए \(16-x^2\ge 0\) चाहिए। इससे \(x^2\le 16\) और डोमेन ([-4,4]) मिलता है। / For a real square root, \(16-x^2\ge 0\) is required. This gives \(x^2\le 16\) and the domain ([-4,4]).
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What is the range of (f(x)=2\sqrt{x+1}-5)?
A. \([-5,\infty\))
\(\sqrt{x+1}\ge 0\), इसलिए \(2\sqrt{x+1}-5\ge -5\)। अतः परिसर \([-5,\infty\)) है। / Since \(\sqrt{x+1}\ge 0\), \(2\sqrt{x+1}-5\ge -5\). Hence the range is \([-5,\infty\)).
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What is the maximum value of (f(x)=3-\sqrt{x-1})?
A. (3)
\(\sqrt{x-1}\ge 0\), इसलिए \(3-\sqrt{x-1}\le 3\)। अधिकतम मान (3) (x=1) पर मिलता है। / Since \(\sqrt{x-1}\ge 0\), \(3-\sqrt{x-1}\le 3\). The maximum value (3) occurs at (x=1).
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What is the starting point of (f(x)=\sqrt{x+6}-2)?
A. ((-6,-2))
\(\sqrt{x+6}\) (x=-6) से शुरू होता है और बाहर (-2) जुड़ता है। इसलिए प्रारंभिक बिंदु ((-6,-2)) है। / \(\sqrt{x+6}\) starts at (x=-6), and (-2) is added outside. So the starting point is ((-6,-2)).
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What is the domain of (f(x)=\sqrt{2x+6})?
A. \(x\ge -3\)
वास्तविक वर्गमूल के लिए \(2x+6\ge 0\) चाहिए। इससे \(x\ge -3\) मिलता है। / For a real square root, \(2x+6\ge 0\) is required. This gives \(x\ge -3\).
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What is the range of (f(x)=\sqrt{5-x}+2)?
A. \([2,\infty\))
\(\sqrt{5-x}\ge 0\), इसलिए (f(x)\ge 2)। (x) घटाने पर वर्गमूल बड़ा हो सकता है, इसलिए परिसर \([2,\infty\)) है। / Since \(\sqrt{5-x}\ge 0\), (f(x)\ge 2). The square root can grow as (x) decreases, so the range is \([2,\infty\)).
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What is the minimum value of (f(x)=2\sqrt{x-1}+3)?
A. (3)
\(\sqrt{x-1}\ge 0\), इसलिए \(2\sqrt{x-1}+3\ge 3\)। न्यूनतम मान (3) (x=1) पर मिलता है। / Since \(\sqrt{x-1}\ge 0\), \(2\sqrt{x-1}+3\ge 3\). The minimum value (3) occurs at (x=1).
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What is the starting point of (f(x)=\sqrt{x+2}-3)?
A. ((-2,-3))
\(\sqrt{x+2}\) तब शुरू होता है जब (x+2=0), अर्थात (x=-2)। बाहर (-3) होने से प्रारंभिक बिंदु ((-2,-3)) है। / \(\sqrt{x+2}\) starts when (x+2=0), that is (x=-2). The outside (-3) makes the starting point ((-2,-3)).
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What is the domain of (f(x)=\sqrt{7-x})?
A. \(x\le 7\)
वास्तविक मान के लिए \(7-x\ge 0\), इसलिए \(x\le 7\)। वर्गमूल में अंदर की राशि ऋणात्मक नहीं होनी चाहिए। / For real values, \(7-x\ge 0\), so \(x\le 7\). The expression inside a square root must not be negative.
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What is the domain of (f(x)=\sqrt{x-4}+1)?
A. \(x\ge 4\)
वास्तविक वर्गमूल के लिए \(x-4\ge 0\) चाहिए। इसलिए डोमेन \(x\ge 4\) है। / For a real square root, \(x-4\ge 0\) is required. Hence the domain is \(x\ge 4\).
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