3.7 Continuity And Differentiablityap Calculus

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3.7 Continuity And Differentiablityap Calculus Algebra

Calculus Introduction: Continuity and Differentiability Notes, Examples, and Practice Quiz (w/solutions) Topics include definition of continuous, limits and asymptotes, differentiable function, and more. Worksheet 3:7 Continuity and Limits Section 1Limits Limits were mentioned without very much explanation in the previous worksheet. We will now take a closer look at limits and, in particular, the limits of functions. Limits are very important in maths, but more specically in calculus.

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Section 2-9 : Continuity

  1. The graph of (fleft( x right)) is given below. Based on this graph determine where the function is discontinuous. Solution
  2. The graph of (fleft( x right)) is given below. Based on this graph determine where the function is discontinuous. Solution
3.7 Continuity And Differentiablityap Calculus

For problems 3 – 7 using only Properties 1 – 9 from the Limit Properties section, one-sided limit properties (if needed) and the definition of continuity determine if the given function is continuous or discontinuous at the indicated points.

  1. (displaystyle fleft( x right) = frac{{4x + 5}}{{9 - 3x}})
    1. (x = - 1)
    2. (x = 0)
    3. (x = 3)
    Solution
  2. (displaystyle gleft( z right) = frac{6}{{{z^2} - 3z - 10}})
    1. (z = - 2)
    2. (z = 0)
    3. (z = 5)
    Solution
  3. (gleft( x right) = left{ {begin{array}{rl}{2x}&{x < 6}{x - 1}&{x ge 6}end{array}} right.)
    1. (x = 4)
    2. (x = 6)
    Solution
  4. (hleft( t right) = left{ {begin{array}{rl}{{t^2}}&{t < - 2}{t + 6}&{t ge - 2}end{array}} right.)
    1. (t = - 2)
    2. (t = 10)
    Solution
  5. (gleft( 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)
    Solution

For problems 8 – 12 determine where the given function is discontinuous.

3.7 Continuity And Differentiablityap Calculus Calculator

3.7 Continuity And Differentiablityap Calculus
  1. (displaystyle fleft( x right) = frac{{{x^2} - 9}}{{3{x^2} + 2x - 8}}) Solution
  2. (displaystyle Rleft( t right) = frac{{8t}}{{{t^2} - 9t - 1}}) Solution
  3. (displaystyle hleft( z right) = frac{1}{{2 - 4cos left( {3z} right)}}) Solution
  4. (displaystyle yleft( x right) = frac{x}{{7 - {{bf{e}}^{2x + 3}}}}) Solution
  5. (gleft( x right) = tan left( {2x} right)) Solution

3.7 Continuity And Differentiablityap Calculus Solver

For problems 13 – 15 use the Intermediate Value Theorem to show that the given equation has at least one solution in the indicated interval. Note that you are NOT asked to find the solution only show that at least one must exist in the indicated interval.

3.7 Continuity And Differentiablityap Calculus 14th Edition

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