Asosiy tarkibga o'tish

2.9-bo'lim: Davomiylik

  1. \(f\left( x \right)\) grafigi quyida keltirilgan. Ushbu grafik asosida funktsiyaning qayerda uzluksiz ekanligini aniqlang.
    The graph of \(f\left( x \right)\) on \( - 6 \le x \le 5\). Starting at the left the graph increases to a peak at approximately \(\left( { - 5,4} \right)\) and then decreases to a closed dot at \(\left( { - 4,3} \right)\). There is an open dot at \(\left( { - 4, - 2} \right)\) and from there the graph increases to a closed dot at \(\left( { - 1,4} \right)\). The graph continues on from that point up to a peak at approximately \(\left( {1,5} \right)\) and then falls steeply to a closed dot at \(\left( {2, - 1} \right)\). There is an open dot at \(\left( {2,5} \right)\) and from there the graph rises to a peak at approximately \(\left( {3,6} \right)\) and then decreases to an open dot at \(\left( {4,2} \right)\). The graph then continues on down from that point, leaving the bottom of the graph at approximately \(x = 5\).
    Yechim
  2. \(f\left( x \right)\) grafigi quyida keltirilgan. Ushbu grafik asosida funktsiyaning qayerda uzluksiz ekanligini aniqlang.
    The graph of \(f\left( x \right)\) on \( - 12 \le x \le 12\). Coming down from the upper left the graph decreases to an open dot at the bottom of a valley at approximately \(\left( { - 8, - 6} \right)\), with a separate closed dot sitting above it at approximately \(\left( { - 8, - 3} \right)\). From the open dot the graph increases, crossing the x-axis near \(x = - 4\), and rises to a closed dot at approximately \(\left( { - 2,3} \right)\) with the graph continuing up off the top of the picture just to the right of that point. The graph picks back up near \(\left( {0,1} \right)\), rises to a small peak at approximately \(\left( {3,3} \right)\), dips slightly and then runs out to an open dot at \(\left( {6,2} \right)\), with a separate closed dot sitting above it at \(\left( {6,5} \right)\). From the closed dot the graph decreases through a closed dot at approximately \(\left( {10,1} \right)\) and continues on down to the lower right.
    Yechim

3-7-muammolar uchun faqat 1-9-chi xususiyatlardan foydalaniladi Cheklash xususiyatlari bo'lim, bir tomonlama chegara xususiyatlari (agar kerak bo'lsa) va uzluksizlik ta'rifi berilgan funktsiyaning ko'rsatilgan nuqtalarda uzluksiz yoki uzluksiz ekanligini aniqlaydi.

  1. \(\displaystyle f\left( x \right) = \frac{{4x + 5}}{{9 - 3x}}\)
    1. \(x = - 1\)
    2. \(x = 0\)
    3. \(x = 3\)
    Yechim
  2. \(\displaystyle g\left( z \right) = \frac{6}{{{z^2} - 3z - 10}}\)
    1. \(z = - 2\)
    2. \(z = 0\)
    3. \(z = 5\)
    Yechim
  3. \(g\left( x \right) = \left\{ {\begin{array}{rl}{2x}&{x < 6}\\{x - 1}&{x \ge 6}\end{array}} \right.\)
    1. \(x = 4\)
    2. \(x = 6\)
    Yechim
  4. \(h\left( t \right) = \left\{ {\begin{array}{rl}{{t^2}}&{t < - 2}\\{t + 6}&{t \ge - 2}\end{array}} \right.\)
    1. \(t = - 2\)
    2. \(t = 10\)
    Yechim
  5. \(g\left( x \right) = \left\{ {\begin{array}{rc}{1 - 3x}&{x < - 6}\\7&{x = - 6}\\{{x^3}}&{ - 6 < x < 1}\\1&{x = 1}\\{2 - x}&{x > 1}\end{array}} \right.\)
    1. \(x = - 6\)
    2. \(x = 1\)
    Yechim

8-12-masalalar uchun berilgan funksiya qayerda uzluksiz ekanligini aniqlang.

  1. \(\displaystyle f\left( x \right) = \frac{{{x^2} - 9}}{{3{x^2} + 2x - 8}}\) Yechim
  2. \(\displaystyle R\left( t \right) = \frac{{8t}}{{{t^2} - 9t - 1}}\) Yechim
  3. \(\displaystyle h\left( z \right) = \frac{1}{{2 - 4\cos \left( {3z} \right)}}\) Yechim
  4. \(\displaystyle y\left( x \right) = \frac{x}{{7 - {{\bf{e}}^{2x + 3}}}}\) Yechim
  5. \(g\left( x \right) = \tan \left( {2x} \right)\) Yechim

13-15-masalalar uchun Oraliq qiymat teoremasidan foydalanib, berilgan tenglama ko'rsatilgan oraliqda kamida bitta yechimga ega ekanligini ko'rsating. E'tibor bering, sizdan yechim topish so'ralmaydi, faqat ko'rsatilgan intervalda kamida bittasi bo'lishi kerakligini ko'rsating.

  1. \(25 - 8{x^2} - {x^3} = 0\) on \(\left[ { - 2,4} \right]\) Yechim
  2. \({w^2} - 4\ln \left( {5w + 2} \right) = 0\) on \(\left[ {0,4} \right]\) Yechim
  3. \(4t + 10{{\bf{e}}^t} - {{\bf{e}}^{2t}} = 0\) on \(\left[ {1,3} \right]\) Yechim