Bài 1. Sự đồng biến, nghịch biến của hàm số
Lý thuyết Sự đồng biến, nghịch biến của hàm số
<h3><strong>1. H&agrave;m số đồng biến, nghịch biến l&agrave; g&igrave;?</strong></h3> <p>K&iacute; hiệu: K l&agrave; một khoảng, một đoạn hoặc một nửa khoảng.</p> <p>Cho h&agrave;m số&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>y</mi><mo>&#160;</mo><mo>=</mo><mo>&#160;</mo><mi>f</mi><mfenced><mi>x</mi></mfenced></math> x&aacute;c định tr&ecirc;n K</p> <p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mo>&#160;</mo><mi>H</mi><mi>&#224;</mi><mi>m</mi><mo>&#160;</mo><mi>s</mi><mi>&#7889;</mi><mo>&#160;</mo><mi>y</mi><mo>&#160;</mo><mo>=</mo><mo>&#160;</mo><mi>f</mi><mfenced><mi>x</mi></mfenced><mo>&#160;</mo><mi>&#273;</mi><mi>&#7891;</mi><mi>n</mi><mi>g</mi><mo>&#160;</mo><mi>b</mi><mi>i</mi><mi>&#7871;</mi><mi>n</mi><mo>&#160;</mo><mo>(</mo><mi>t</mi><mi>&#259;</mi><mi>n</mi><mi>g</mi><mo>)</mo><mo>&#160;</mo><mi>t</mi><mi>r</mi><mi>&#234;</mi><mi>n</mi><mo>&#160;</mo><mi>K</mi><mo>&#160;</mo><mi>n</mi><mi>&#7871;</mi><mi>u</mi><mo>&#160;</mo><mfenced open="{" close=""><mtable columnalign="left"><mtr><mtd><msub><mi>x</mi><mn>1</mn></msub><mo>,</mo><msub><mi>x</mi><mn>2</mn></msub><mo>&#160;</mo><mo>&#8712;</mo><mo>&#160;</mo><mi>K</mi></mtd></mtr><mtr><mtd><msub><mi>x</mi><mn>1</mn></msub><mo>&#160;</mo><mo>&#60;</mo><mo>&#160;</mo><msub><mi>x</mi><mn>2</mn></msub></mtd></mtr></mtable></mfenced><mo>&#8658;</mo><mo>&#160;</mo><mi>f</mi><mo>(</mo><msub><mi>x</mi><mn>1</mn></msub><mo>)</mo><mo>&#160;</mo><mo>&#60;</mo><mo>&#160;</mo><mi>f</mi><mfenced><msub><mi>x</mi><mn>2</mn></msub></mfenced></math></p> <p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mo>&#160;</mo><mi>H</mi><mi>&#224;</mi><mi>m</mi><mo>&#160;</mo><mi>s</mi><mi>&#7889;</mi><mo>&#160;</mo><mi>y</mi><mo>&#160;</mo><mo>=</mo><mo>&#160;</mo><mi>f</mi><mfenced><mi>x</mi></mfenced><mo>&#160;</mo><mi>n</mi><mi>g</mi><mi>h</mi><mi>&#7883;</mi><mi>c</mi><mi>h</mi><mo>&#160;</mo><mi>b</mi><mi>i</mi><mi>&#7871;</mi><mi>n</mi><mo>&#160;</mo><mo>(</mo><mi>g</mi><mi>i</mi><mi>&#7843;</mi><mi>m</mi><mo>)</mo><mo>&#160;</mo><mi>t</mi><mi>r</mi><mi>&#234;</mi><mi>n</mi><mo>&#160;</mo><mi>K</mi><mo>&#160;</mo><mi>n</mi><mi>&#7871;</mi><mi>u</mi><mo>&#160;</mo><mfenced open="{" close=""><mtable columnalign="left"><mtr><mtd><msub><mi>x</mi><mn>1</mn></msub><mo>,</mo><msub><mi>x</mi><mn>2</mn></msub><mo>&#160;</mo><mo>&#8712;</mo><mo>&#160;</mo><mi>K</mi></mtd></mtr><mtr><mtd><msub><mi>x</mi><mn>1</mn></msub><mo>&#160;</mo><mo>&#60;</mo><mo>&#160;</mo><msub><mi>x</mi><mn>2</mn></msub></mtd></mtr></mtable></mfenced><mo>&#8658;</mo><mo>&#160;</mo><mi>f</mi><mo>(</mo><msub><mi>x</mi><mn>1</mn></msub><mo>)</mo><mo>&#160;</mo><mo>&#62;</mo><mo>&#160;</mo><mi>f</mi><mfenced><msub><mi>x</mi><mn>2</mn></msub></mfenced></math></p> <h3><strong>2. Điều kiện cần để h&agrave;m số đơn điệu</strong></h3> <p>Cho h&agrave;m số y = f(x) c&oacute; đạo h&agrave;m tr&ecirc;n K,</p> <p>- Nếu f(x) đồng biến tr&ecirc;n K th&igrave;&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mo>'</mo><mfenced><mi>x</mi></mfenced><mo>&#160;</mo><mo>&#8805;</mo><mo>&#160;</mo><mn>0</mn><mo>&#160;</mo><mi>v</mi><mi>&#7899;</mi><mi>i</mi><mo>&#160;</mo><mi>m</mi><mi>&#7885;</mi><mi>i</mi><mo>&#160;</mo><mi>x</mi><mo>&#160;</mo><mo>&#8712;</mo><mo>&#160;</mo><mi>K</mi></math></p> <p>- Nếu f(x) nghịch biến tr&ecirc;n K th&igrave; <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mo>'</mo><mfenced><mi>x</mi></mfenced><mo>&#160;</mo><mo>&#8804;</mo><mo>&#160;</mo><mn>0</mn><mo>&#160;</mo><mi>v</mi><mi>&#7899;</mi><mi>i</mi><mo>&#160;</mo><mi>m</mi><mi>&#7885;</mi><mi>i</mi><mo>&#160;</mo><mi>x</mi><mo>&#160;</mo><mo>&#8712;</mo><mo>&#160;</mo><mi>K</mi></math></p> <h3><strong>3. Điều kiện đủ để h&agrave;m số đơn điệu</strong></h3> <p>Cho h&agrave;m số y = f(x) c&oacute; đạo h&agrave;m tr&ecirc;n K:&nbsp;</p> <p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mo>&#160;</mo><mi>N</mi><mi>&#7871;</mi><mi>u</mi><mo>&#160;</mo><mi>f</mi><mo>'</mo><mfenced><mi>x</mi></mfenced><mo>&#160;</mo><mo>&#8805;</mo><mo>&#160;</mo><mn>0</mn><mo>&#160;</mo><mi>v</mi><mi>&#7899;</mi><mi>i</mi><mo>&#160;</mo><mi>m</mi><mi>&#7885;</mi><mi>i</mi><mo>&#160;</mo><mi>x</mi><mo>&#160;</mo><mo>&#8712;</mo><mo>&#160;</mo><mi>K</mi><mo>&#160;</mo><mi>v</mi><mi>&#224;</mi><mo>&#160;</mo><mi>f</mi><mo>'</mo><mfenced><mi>x</mi></mfenced><mo>&#160;</mo><mo>=</mo><mo>&#160;</mo><mn>0</mn><mo>&#160;</mo><mo>&#160;</mo><mi>c</mi><mi>h</mi><mi>&#7881;</mi><mo>&#160;</mo><mi>t</mi><mi>&#7841;</mi><mi>i</mi><mo>&#160;</mo><mi>m</mi><mi>&#7897;</mi><mi>t</mi><mo>&#160;</mo><mi>s</mi><mi>&#7889;</mi><mo>&#160;</mo><mi>h</mi><mi>&#7919;</mi><mi>u</mi><mo>&#160;</mo><mi>h</mi><mi>&#7841;</mi><mi>n</mi><mo>&#160;</mo><mi>&#273;</mi><mi>i</mi><mi>&#7875;</mi><mi>m</mi><mo>&#160;</mo><mi>t</mi><mi>h</mi><mi>u</mi><mi>&#7897;</mi><mi>c</mi><mo>&#160;</mo><mi>K</mi><mo>&#160;</mo><mi>t</mi><mi>h</mi><mi>&#236;</mi><mo>&#160;</mo><mi>f</mi><mfenced><mi>x</mi></mfenced><mo>&#160;</mo><mi>&#273;</mi><mi>&#7891;</mi><mi>n</mi><mi>g</mi><mo>&#160;</mo><mi>b</mi><mi>i</mi><mi>&#7871;</mi><mi>n</mi><mo>&#160;</mo><mi>t</mi><mi>r</mi><mi>&#234;</mi><mi>n</mi><mo>&#160;</mo><mi>K</mi><mspace linebreak="newline"/><mo>-</mo><mo>&#160;</mo><mi>N</mi><mi>&#7871;</mi><mi>u</mi><mo>&#160;</mo><mi>f</mi><mo>'</mo><mfenced><mi>x</mi></mfenced><mo>&#160;</mo><mo>&#8804;</mo><mo>&#160;</mo><mn>0</mn><mo>&#160;</mo><mi>v</mi><mi>&#7899;</mi><mi>i</mi><mo>&#160;</mo><mi>m</mi><mi>&#7885;</mi><mi>i</mi><mo>&#160;</mo><mi>x</mi><mo>&#160;</mo><mo>&#8712;</mo><mo>&#160;</mo><mi>K</mi><mo>&#160;</mo><mi>v</mi><mi>&#224;</mi><mo>&#160;</mo><mi>f</mi><mo>'</mo><mfenced><mi>x</mi></mfenced><mo>&#160;</mo><mo>=</mo><mo>&#160;</mo><mn>0</mn><mo>&#160;</mo><mo>&#160;</mo><mi>c</mi><mi>h</mi><mi>&#7881;</mi><mo>&#160;</mo><mi>t</mi><mi>&#7841;</mi><mi>i</mi><mo>&#160;</mo><mi>m</mi><mi>&#7897;</mi><mi>t</mi><mo>&#160;</mo><mi>s</mi><mi>&#7889;</mi><mo>&#160;</mo><mi>h</mi><mi>&#7919;</mi><mi>u</mi><mo>&#160;</mo><mi>h</mi><mi>&#7841;</mi><mi>n</mi><mo>&#160;</mo><mi>&#273;</mi><mi>i</mi><mi>&#7875;</mi><mi>m</mi><mo>&#160;</mo><mi>t</mi><mi>h</mi><mi>u</mi><mi>&#7897;</mi><mi>c</mi><mo>&#160;</mo><mi>K</mi><mo>&#160;</mo><mi>t</mi><mi>h</mi><mi>&#236;</mi><mo>&#160;</mo><mi>f</mi><mfenced><mi>x</mi></mfenced><mo>&#160;</mo><mi>n</mi><mi>g</mi><mi>h</mi><mi>&#7883;</mi><mi>c</mi><mi>h</mi><mo>&#160;</mo><mi>b</mi><mi>i</mi><mi>&#7871;</mi><mi>n</mi><mo>&#160;</mo><mi>t</mi><mi>r</mi><mi>&#234;</mi><mi>n</mi><mo>&#160;</mo><mi>K</mi><mspace linebreak="newline"/><mo>-</mo><mo>&#160;</mo><mi>N</mi><mi>&#7871;</mi><mi>u</mi><mo>&#160;</mo><mi>f</mi><mo>'</mo><mfenced><mi>x</mi></mfenced><mo>&#160;</mo><mo>=</mo><mo>&#160;</mo><mn>0</mn><mo>&#160;</mo><mi>v</mi><mi>&#7899;</mi><mi>i</mi><mo>&#160;</mo><mi>m</mi><mi>&#7885;</mi><mi>i</mi><mo>&#160;</mo><mi>x</mi><mo>&#160;</mo><mo>&#8712;</mo><mo>&#160;</mo><mi>K</mi><mo>&#160;</mo><mi>t</mi><mi>h</mi><mi>&#236;</mi><mo>&#160;</mo><mi>f</mi><mfenced><mi>x</mi></mfenced><mo>&#160;</mo><mi>l</mi><mi>&#224;</mi><mo>&#160;</mo><mi>h</mi><mi>&#224;</mi><mi>m</mi><mo>&#160;</mo><mi>h</mi><mi>&#7857;</mi><mi>n</mi><mi>g</mi><mo>&#160;</mo><mi>t</mi><mi>r</mi><mi>&#234;</mi><mi>n</mi><mo>&#160;</mo><mi>K</mi></math></p> <h3><strong>4. C&aacute;c bước x&eacute;t t&iacute;nh đơn điệu của h&agrave;m số</strong></h3> <p>Để x&eacute;t t&iacute;nh đơn điệu của h&agrave;m số, ta thực hiện theo c&aacute;c bước sau:</p> <p><strong>- Bước</strong> <strong>1:</strong> T&igrave;m tập x&aacute;c định của h&agrave;m số</p> <p><strong>- Bước 2:</strong> T&iacute;nh đạo h&agrave;m f'(x) = 0 v&agrave; t&igrave;m c&aacute;c điểm x<sub>i</sub> (với i = 1, 2, ..., n)&nbsp;m&agrave; tại đ&oacute; đạo h&agrave;m bằng 0 hoặc kh&ocirc;ng x&aacute;c định</p> <p><strong>- Bước 3:</strong> Sắp xếp c&aacute;c điểm x<sub>i</sub>&nbsp;theo thứ tự tăng dần v&agrave; lập bảng biến thi&ecirc;n</p> <p><strong>- Bước 4:</strong> N&ecirc;u kết luận về c&aacute;c khoảng đồng biến, nghịch biến của h&agrave;m số.</p>
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