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*3.7 Continuity And Differentiablityap Calculus Calculator
*3.7 Continuity And Differentiablityap Calculus Solver
*3.7 Continuity And Differentiablityap Calculus 14th EditionShow Mobile NoticeShow All NotesHide All NotesYou appear to be on a device with a ’narrow’ screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.Section 2-9 : Continuity
Math AP®︎/College Calculus AB Differentiation: definition and basic derivative rules Connecting differentiability and continuity: determining when derivatives do and do not exist Differentiability and continuity. Consider the step function f which returns a value, say 1, for all x less than a, and returns a different value, say 5, for all x greater than or equal to a.This function does not have a derivative at the marked point, as the function is not continuous there.
*The graph of (fleft( x right)) is given below. Based on this graph determine where the function is discontinuous. Solution
*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.
*(displaystyle fleft( x right) = frac{{4x + 5}}{{9 - 3x}})
*(x = - 1)
*(x = 0)
*(x = 3)Solution
*(displaystyle gleft( z right) = frac{6}{{{z^2} - 3z - 10}})
*(z = - 2)
*(z = 0)
*(z = 5)Solution
*(gleft( x right) = left{ {begin{array}{rl}{2x}&{x < 6}{x - 1}&{x ge 6}end{array}} right.)
*(x = 4)
*(x = 6)Solution
*(hleft( t right) = left{ {begin{array}{rl}{{t^2}}&{t < - 2}{t + 6}&{t ge - 2}end{array}} right.)
*(t = - 2)
*(t = 10)Solution
*(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.)
*(x = - 6)
*(x = 1)Solution3.7 Continuity And Differentiablityap Calculus Calculator
For problems 8 – 12 determine where the given function is discontinuous.
*(displaystyle fleft( x right) = frac{{{x^2} - 9}}{{3{x^2} + 2x - 8}}) Solution
*(displaystyle Rleft( t right) = frac{{8t}}{{{t^2} - 9t - 1}}) Solution
*(displaystyle hleft( z right) = frac{1}{{2 - 4cos left( {3z} right)}}) Solution
*(displaystyle yleft( x right) = frac{x}{{7 - {{bf{e}}^{2x + 3}}}}) Solution
*(gleft( x right) = tan left( {2x} right)) Solution
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.
*(25 - 8{x^2} - {x^3} = 0) on (left[ { - 2,4} right]) Solution
*({w^2} - 4ln left( {5w + 2} right) = 0) on (left[ {0,4} right]) Solution
*(4t + 10{{bf{e}}^t} - {{bf{e}}^{2t}} = 0) on (left[ {1,3} right]) SolutionShow Mobile NoticeShow All NotesHide All NotesYou appear to be on a device with a ’narrow’ screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.Section 2-9 : Continuity
*The graph of (fleft( x right)) is given below. Based on this graph determine where the function is discontinuous. Solution
*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.
*(displaystyle fleft( x right) = frac{{4x + 5}}{{9 - 3x}})
*(x = - 1)
*(x = 0)
*(x = 3)Solution
*(displaystyle gleft( z right) = frac{6}{{{z^2} - 3z - 10}})
*(z = - 2)
*(z = 0)
*(z = 5)Solution
*(gleft( x right) = left{ {begin{array}{rl}{2x}&{x < 6}{x - 1}&{x ge 6}end{array}} right.)
*(x = 4)
*(x = 6)Solution
*(hleft( t right) = left{ {begin{array}{rl}{{t^2}}&{t < - 2}{t + 6}&{t ge - 2}end{array}} right.)
*(t = - 2)
*(t = 10)Solution
*(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.)
*(x = - 6)
*(x = 1)Solution3.7 Continuity And Differentiablityap Calculus Solver
For problems 8 – 12 determine where the given function is discontinuous.
*(displaystyle fleft( x right) = frac{{{x^2} - 9}}{{3{x^2} + 2x - 8}}) Solution
*(displaystyle Rleft( t right) = frac{{8t}}{{{t^2} - 9t - 1}}) Solution
*(displaystyle hleft( z right) = frac{1}{{2 - 4cos left( {3z} right)}}) Solution
*(displaystyle yleft( x right) = frac{x}{{7 - {{bf{e}}^{2x + 3}}}}) Solution
*(gleft( x right) = tan left( {2x} right)) Solution3.7 Continuity And Differentiablityap Calculus 14th Edition
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.
*(25 - 8{x^2} - {x^3} = 0) on (left[ { - 2,4} right]) Solution
*({w^2} - 4ln left( {5w + 2} right) = 0) on (left[ {0,4} right]) Solution
*(4t + 10{{bf{e}}^t} - {{bf{e}}^{2t}} = 0) on (left[ {1,3} right]) Solution
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*3.7 Continuity And Differentiablityap Calculus Calculator
*3.7 Continuity And Differentiablityap Calculus Solver
*3.7 Continuity And Differentiablityap Calculus 14th EditionShow Mobile NoticeShow All NotesHide All NotesYou appear to be on a device with a ’narrow’ screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.Section 2-9 : Continuity
Math AP®︎/College Calculus AB Differentiation: definition and basic derivative rules Connecting differentiability and continuity: determining when derivatives do and do not exist Differentiability and continuity. Consider the step function f which returns a value, say 1, for all x less than a, and returns a different value, say 5, for all x greater than or equal to a.This function does not have a derivative at the marked point, as the function is not continuous there.
*The graph of (fleft( x right)) is given below. Based on this graph determine where the function is discontinuous. Solution
*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.
*(displaystyle fleft( x right) = frac{{4x + 5}}{{9 - 3x}})
*(x = - 1)
*(x = 0)
*(x = 3)Solution
*(displaystyle gleft( z right) = frac{6}{{{z^2} - 3z - 10}})
*(z = - 2)
*(z = 0)
*(z = 5)Solution
*(gleft( x right) = left{ {begin{array}{rl}{2x}&{x < 6}{x - 1}&{x ge 6}end{array}} right.)
*(x = 4)
*(x = 6)Solution
*(hleft( t right) = left{ {begin{array}{rl}{{t^2}}&{t < - 2}{t + 6}&{t ge - 2}end{array}} right.)
*(t = - 2)
*(t = 10)Solution
*(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.)
*(x = - 6)
*(x = 1)Solution3.7 Continuity And Differentiablityap Calculus Calculator
For problems 8 – 12 determine where the given function is discontinuous.
*(displaystyle fleft( x right) = frac{{{x^2} - 9}}{{3{x^2} + 2x - 8}}) Solution
*(displaystyle Rleft( t right) = frac{{8t}}{{{t^2} - 9t - 1}}) Solution
*(displaystyle hleft( z right) = frac{1}{{2 - 4cos left( {3z} right)}}) Solution
*(displaystyle yleft( x right) = frac{x}{{7 - {{bf{e}}^{2x + 3}}}}) Solution
*(gleft( x right) = tan left( {2x} right)) Solution
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.
*(25 - 8{x^2} - {x^3} = 0) on (left[ { - 2,4} right]) Solution
*({w^2} - 4ln left( {5w + 2} right) = 0) on (left[ {0,4} right]) Solution
*(4t + 10{{bf{e}}^t} - {{bf{e}}^{2t}} = 0) on (left[ {1,3} right]) SolutionShow Mobile NoticeShow All NotesHide All NotesYou appear to be on a device with a ’narrow’ screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.Section 2-9 : Continuity
*The graph of (fleft( x right)) is given below. Based on this graph determine where the function is discontinuous. Solution
*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.
*(displaystyle fleft( x right) = frac{{4x + 5}}{{9 - 3x}})
*(x = - 1)
*(x = 0)
*(x = 3)Solution
*(displaystyle gleft( z right) = frac{6}{{{z^2} - 3z - 10}})
*(z = - 2)
*(z = 0)
*(z = 5)Solution
*(gleft( x right) = left{ {begin{array}{rl}{2x}&{x < 6}{x - 1}&{x ge 6}end{array}} right.)
*(x = 4)
*(x = 6)Solution
*(hleft( t right) = left{ {begin{array}{rl}{{t^2}}&{t < - 2}{t + 6}&{t ge - 2}end{array}} right.)
*(t = - 2)
*(t = 10)Solution
*(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.)
*(x = - 6)
*(x = 1)Solution3.7 Continuity And Differentiablityap Calculus Solver
For problems 8 – 12 determine where the given function is discontinuous.
*(displaystyle fleft( x right) = frac{{{x^2} - 9}}{{3{x^2} + 2x - 8}}) Solution
*(displaystyle Rleft( t right) = frac{{8t}}{{{t^2} - 9t - 1}}) Solution
*(displaystyle hleft( z right) = frac{1}{{2 - 4cos left( {3z} right)}}) Solution
*(displaystyle yleft( x right) = frac{x}{{7 - {{bf{e}}^{2x + 3}}}}) Solution
*(gleft( x right) = tan left( {2x} right)) Solution3.7 Continuity And Differentiablityap Calculus 14th Edition
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.
*(25 - 8{x^2} - {x^3} = 0) on (left[ { - 2,4} right]) Solution
*({w^2} - 4ln left( {5w + 2} right) = 0) on (left[ {0,4} right]) Solution
*(4t + 10{{bf{e}}^t} - {{bf{e}}^{2t}} = 0) on (left[ {1,3} right]) Solution
Download here: http://gg.gg/o9obm
https://diarynote-jp.indered.space
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