10/7/2021

3.7 Continuity And Differentiablityap Calculus

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This calculus video tutorial provides a basic introduction into continuity and differentiability. Continuity tells you if the function f(x) is continuous. 3.7 Continuity and Differentiablity Notes 3.7 Key. Powered by Create your own unique website with customizable templates.

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Section 2-9 : Continuity

  • This follows from the difference-quotient definition of the derivative. The last equality follows from the continuity of the derivatives at c. The limit in the conclusion is not indeterminate because. Lim x→0 sinx x!!!!!
  • Continuity in Calculus: Definition, Examples & Problems. If I round 3.7, I get 4. Well here, if I graph f(x) as a function of x, I have to pick my finger up. In fact, I need to pick my finger.
3.7 continuity and differentiablityap calculus 14th edition
  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

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.

3.7 Continuity And Differentiablityap Calculus Solver

  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
  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 Algebra

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.

Solver
  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