diff --git a/source/calculus/source/01-LT/04.ptx b/source/calculus/source/01-LT/04.ptx index 8ce2db22f..326a3a334 100644 --- a/source/calculus/source/01-LT/04.ptx +++ b/source/calculus/source/01-LT/04.ptx @@ -43,8 +43,13 @@

- + + +

+ C. The volume of water in a tank that is gradually filled over time +

+ @@ -84,29 +89,71 @@ + +

For each of the values a = -3, -2, -1, 0, 1, 2, 3, determine whether the limit \displaystyle\lim_{x \to a} f(x) exists. If the limit does not exist, be ready to explain why not.

+ + +

+ Limit exists at a= -3, -1 0, 1, 3 +

+

+ Limit does not exist at a= -2 (Limit from right is not equal to limit from left) and a= 2 ( the function oscillates at a= 2 ) +

+ +

For each of the values of a where the limit of f exists, determine the value of f(a) at each such point.

+ + +

+ f(-3) = 3 , f(-1)= 1 , f(0)= -4 , f(1)= -2.5 and f(3) DNE. +

+ +

For each such a value, is f(a) equal to \displaystyle\lim_{x \to a} f(x)?

+ + +

+ f(-3)= 3 and \displaystyle\lim_{x \to -3} f(x)= 3 +

+

+ f(0)= -4 and \displaystyle\lim_{x \to 0} f(x)= -4 +

+ +

Use your understanding of continuity to determine whether f is continuous at each value of a.

+ + +

+ f is continuous at those points where the value of function is equal to the limit of the function at the point. +

+ +

Are there any revisions you would make to the definition of continuity that you arrived at toward the end of ?

+ + +

+ f is continuous if at a point a if \displaystyle\lim_{x \to a} f(x)= f(a) +

+ @@ -145,6 +192,7 @@

\displaystyle\lim_{x \to -3^+} h(x)

+

  • @@ -162,7 +210,13 @@

    • + + +

    + All four choices are equal. +

    +     @@ -199,21 +253,44 @@

    For which values of a do we have \displaystyle\lim_{x \to a^-} f(x) \ne \lim_{x \to a^+} f(x)?

    + +

    + At a = -2 and a = 2 +

    +

    For which values of a is f(a) not defined?

    + +

    + At a = 0 , a= 2 and a = 3 +

    +

    For which values of a does f have a limit at a, yet \displaystyle f(a) \ne \lim_{x \to a} f(x)?

    + +

    + At a = -2 , a = -1 , a = 2 , and a = 3 +

    +

    For which values of a does f fail to be continuous? Give a complete list of intervals on which f is continuous.

    + +

    + f is not continuous at a = -2 , a = -1 , a = 2 , and a = 3 +

    +

    + f is continuous at ( -\infty, -2) , (-2,-1) , (-1,2) , (2,3 ) , and (3,\infty) +

    +     @@ -234,7 +311,14 @@

    + + +

    + B. f is continuous at x = a +

    + +     @@ -265,6 +349,11 @@ + +

    + C. Graphs and formulas only +

    +     @@ -297,6 +386,11 @@ Give a list of x-values where f(x) is not continuous. Be prepared to defend your answer based on .

    + +

    + x= 1, 5 and x \geq 7 +

    +     @@ -324,6 +418,11 @@ \end{cases}

    + + +

    To make h(x) continuous at x=5, b=4.

    + +

    @@ -338,17 +437,15 @@ the function f(x).

    \end{cases}

    + + +

    The function f(x) has a jump discontinuity.

    + + +     - + @@ -376,6 +473,11 @@ jump discontinuity. \end{cases}

    + +

    + c = 2 +

    +

    @@ -386,6 +488,11 @@ jump discontinuity. \end{cases}

    + +

    + c = \pm 2 +

    +

    @@ -396,6 +503,11 @@ jump discontinuity. \end{cases}

    + +

    + c = 0 and c = 1 +

    +     @@ -429,11 +541,25 @@ jump discontinuity.

    The part of the theorem that starts with “Suppose…” forms the assumptions of the theorem, while the part of the theorem that starts with “Then…” is the conclusion of the theorem. What are the assumptions of the Intermediate Value Theorem? What is the conclusion? -

         +

    + +

    + Assumption:

    The function f is continuous on the interval [a,b] ;

    +

    If N is the value between f(a) and f(b) such that f(a)\leq N \leq f(b) or f(b)\leq N \leq f(a).

    + + Conclusion: There exists at least a number c within a to b such that f(c) = k +

    +     +

    Apply the Intermediate Value Theorem to show that the function f(x) = x^3 +x -3 has a zero (so crosses the x-axis) at some point between x=-1 and x=2. (Hint: What interval of x values is being considered here? What is N? Why is N between f(a) and f(b)?)

    + +

    + f(-1) = -5 and f(2)= 7 and N = 0 , since -5 \leq 0 \leq 7 +

    +