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.
    This graph consists of four segments. The first is in the domain \(x < -3\). This segment is a decreasing function that starts at (-4,5) and ends at an open dot at (-3,2). There is also a closed dot at (-3,5). The second segment is in the domain \(-3 < x \le -1\). It starts with an open dot at (-3,2) and increases until approximately (-2.2, 4.8) and then decreases until it ends at a closed dot at (-1,4). The third segment is in the domain \(-1 < x < \le 2\). It starts with a closed dot at (-1,-3) has an open dot at (1,-1) and ends at a closed dot at (2,0). The final segment is in the domain \(x > 2\). The graph in the segment is an oscillating function that oscillates faster and faster as it approaches x = 2 from the right and the oscillation slows down as it moves away from x=2.
  2. \(f\left( x \right)\) grafigi quyida keltirilgan. Ushbu grafik asosida funktsiyaning qayerda uzluksiz ekanligini aniqlang.
    This function has three parts to it. In the domain \(-4 \le x <-1\) it is a parabola with a vertex at approximately (-2.2, 4.1) that opens down intersecting the x-axis at approximately (-3.8,0) and ends at (-1,1) in an open dot. Also in this portion there is an open dot at (-2,4) and a closed dot at (-2,2). In the domain \(-1 \le x \le 2\) it is a parabola with vertex at approximately (0.25, 5) that opens upwards. The left end is a closed dot at (-1,-3) and the right end is a closed dot at (2,-1). The final portion is in the domain \(2 < x \le 4\) and is a parabola with vertex at (3,5) which is indicated by a closed dot and opens downward. The left end of the parabola is an open dot at (2,2).
  3. \(f\left( x \right)\) grafigi quyida keltirilgan. Ushbu grafik asosida funktsiyaning qayerda uzluksiz ekanligini aniqlang.

    This function has three parts to it. In the domain \(-5 \le x < 1\) it is a curve that looks pretty much like a parabola with vertex at approximately (-2.5, 3.2) and opens downward. There is a closed dot at (-4,-2), and open dot at (-2,3), a closed dot at (-2,5) and the curve ends at an open dot at (1,-3). The second portion of this graph is in the domain \(1 < x < 4\). It starts with an open dot at (1,4), there is a closed dot at (1,2). As the graph moves to the right it starts out fairly flat with a slight decrease but as it gets closer and closer to x=4 from the left the graph decreases faster and faster. The final portion of the graph is in the domain \(4 < x < 6\). As the graph approaches x=4 from the right it increases faster and faster and as it moves away from x=4 the graph decreases until it reaches the end of the domain at z=6.

4-13-muammolar uchun faqat 1-9-sonli 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. \(\displaystyle f\left( x \right) = \frac{{6 + 2x}}{{7x - 14}}\)
    1. \(x = - 3\)
    2. \(x = 0\)
    3. \(x = 2\)
  2. \(\displaystyle R\left( y \right) = \frac{{2y}}{{{y^2} - 25}}\)
    1. \(y = - 5\)
    2. \(y = - 1\)
    3. \(y = 3\)
  3. \(\displaystyle g\left( z \right) = \frac{{5z - 20}}{{{z^2} - 12z}}\)
    1. \(z = - 1\)
    2. \(z = 0\)
    3. \(z = 4\)
  4. \(\displaystyle W\left( x \right) = \frac{{2 + x}}{{{x^2} + 6x - 7}}\)
    1. \(x = - 7\)
    2. \(x = 0\)
    3. \(x = 1\)
  5. \(h\left( z \right) = \left\{ {\begin{array}{rl}{2{z^2}}&{z < - 1}\\{4z + 6}&{z \ge - 1}\end{array}} \right.\)
    1. \(z = - 6\)
    2. \(z = - 1\)
  6. \(g\left( x \right) = \left\{ {\begin{array}{rl}{x + {{\bf{e}}^x}}&{x < 0}\\{{x^2}}&{x \ge 0}\end{array}} \right.\)
    1. \(x = 0\)
    2. \(x = 4\)
  7. \(Z\left( t \right) = \left\{ {\begin{array}{rl}8&{t < 5}\\{1 - 6t}&{t \ge 5}\end{array}} \right.\)
    1. \(t = 0\)
    2. \(t = 5\)
  8. \(h\left( z \right) = \left\{ {\begin{array}{rl}{z + 2}&{z < - 4}\\0&{z = - 4}\\{18 - {z^2}}&{z > - 4}\end{array}} \right.\)
    1. \(z = - 4\)
    2. \(z = 2\)
  9. \(f\left( x \right) = \left\{ {\begin{array}{rc}{1 - {x^2}}&{x < 2}\\{ - 3}&{x = 2}\\{2x - 7}&{2 < x < 7}\\0&{x = 7}\\{{x^2}}&{x > 7}\end{array}} \right.\)
    1. \(x = 2\)
    2. \(x = 7\)
  10. \(g\left( w \right) = \left\{ {\begin{array}{rc}{3w}&{w < 0}\\0&{w = 0}\\{w + 6}&{0 < w < 8}\\{14}&{w = 8}\\{22 - w}&{w > 8}\end{array}} \right.\)
    1. \(w = 0\)
    2. \(w = 8\)

14-22-masalalar uchun berilgan funksiya qayerda uzluksiz ekanligini aniqlang.

  1. \(\displaystyle f\left( x \right) = \frac{{11 - 2x}}{{2{x^2} - 13x - 7}}\)
  2. \(\displaystyle Q\left( z \right) = \frac{3}{{2{z^2} + 3z - 4}}\)
  3. \(\displaystyle h\left( t \right) = \frac{{{t^2} - 1}}{{{t^3} + 6{t^2} + t}}\)
  4. \(\displaystyle f\left( z \right) = \frac{{4z + 1}}{{5\cos \left( {{\textstyle{z \over 2}}} \right) + 1}}\)
  5. \(\displaystyle h\left( x \right) = \frac{{1 - x}}{{x\sin \left( {x - 1} \right)}}\)
  6. \(\displaystyle f\left( x \right) = \frac{3}{{4{{\bf{e}}^{x - 7}} - 1}}\)
  7. \(\displaystyle R\left( w \right) = \frac{{{{\bf{e}}^{{w^2} + 1}}}}{{{{\bf{e}}^w} - 2{{\bf{e}}^{1 - w}}}}\)
  8. \(g\left( x \right) = \cot \left( {4x} \right)\)
  9. \(f\left( t \right) = \sec \left( {\sqrt t } \right)\)

23-27-masalalar uchun Oraliq qiymat teoremasidan foydalanib, berilgan tenglamaning koʻrsatilgan oraliqda kamida bitta yechimi borligini koʻrsating. E'tibor bering, sizdan yechim topish so'ralmaydi, faqat ko'rsatilgan intervalda kamida bittasi bo'lishi kerakligini ko'rsating.

  1. \(1 + 7{x^3} - {x^4} = 0\), \(\left[ {4,8} \right]\)
  2. \({z^2} + 11z = 3\), \(\left[ { - 15, - 5} \right]\)
  3. \(\displaystyle \frac{{{t^2} + t - 15}}{{t - 8}} = 0\), \(\left[ { - 5,1} \right]\)
  4. \(\ln \left( {2{t^2} + 1} \right) - \ln \left( {{t^2} + 4} \right) = 0\), \(\left[ { - 1,2} \right]\)
  5. \(\left[ {0,4} \right]\) da \(10 = {w^3} + {w^2}{{\bf{e}}^{ - w}} - 5\)

28 - 33 muammolar uchun \(f\left( x \right)\) hamma joyda uzluksiz deb faraz qiling, agar boshqacha ko'rsatilmagan bo'lsa. Berilgan ma'lumotlardan ma'lum intervalda \(f\left( x \right)\) ning ildizi mavjudligini aniqlash mumkinmi?

Agar berilgan oraliqda ildiz borligini aniqlash mumkin bo'lsa, ildiz mavjudligini qanday bilishingizni aniq tushuntiring. Agar oraliqda ildiz borligini aniqlashning iloji bo'lmasa, har biri berilgan ma'lumotlarga mos keladigan ikkita funktsiya grafigini chizing va ulardan biri berilgan oraliqda ildizga ega bo'ladi, ikkinchisi esa berilgan intervalda ildizga ega bo'lmaydi.

  1. \(f\left( { - 5} \right) = 12\) va \(f\left( 0 \right) = - 3\) \(\left[ { - 5,0} \right]\) oralig'ida.
  2. \(f\left( 1 \right) = 30\) va \(f\left( 9 \right) = 6\) \(\left[ {1,9} \right]\) oralig'ida.
  3. \(\left[ {20,40} \right]\) oralig'ida \(f\left( {20} \right) = - 100\) va \(f\left( {40} \right) = - 100\).
  4. \(\left[ { - 4,5} \right]\) oralig'ida \(f\left( { - 4} \right) = - 10\), \(f\left( 5 \right) = 17\), \(\mathop {\lim }\limits_{x \to \,{1^ - }} f\left( x \right) = - 2\) va \(\mathop {\lim }\limits_{x \to \,{1^ + }} f\left( x \right) = 4\).
  5. \(\left[ { - 8,1} \right]\) oralig'ida \(f\left( { - 8} \right) = 2\), \(f\left( 1 \right) = 23\), \(\mathop {\lim }\limits_{x \to \, - {4^ - }} f\left( x \right) = 35\) va \(\mathop {\lim }\limits_{x \to \, - {4^ + }} f\left( x \right) = 1\).
  6. \(f\left( 0 \right) = - 1\), \(f\left( 9 \right) = 10\), \(\mathop {\lim }\limits_{x \to \,{2^ - }} f\left( x \right) = - 12\) va \(\mathop {\lim }\limits_{x \to \,{2^ + }} f\left( x \right) = - 3\) \(\left[ {0,9} \right]\) oralig'ida.