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2 x 2 5 x + 3 x 1 at x = 1

No. It is a removable discontinuity.

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h ( θ ) = sin θ cos θ tan θ at θ = π

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g ( u ) = { 6 u 2 + u 2 2 u 1 if u 1 2 7 2 if u = 1 2 , at u = 1 2

Yes. It is continuous.

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f ( y ) = sin ( π y ) tan ( π y ) , at y = 1

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f ( x ) = { x 2 e x if x < 0 x 1 if x 0 , at x = 0

Yes. It is continuous.

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f ( x ) = { x sin ( x ) if x π x tan ( x ) if x > π , at x = π

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In the following exercises, find the value(s) of k that makes each function continuous over the given interval.

f ( x ) = { 3 x + 2 , x < k 2 x 3 , k x 8

k = −5

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f ( θ ) = { sin θ , 0 θ < π 2 cos ( θ + k ) , π 2 θ π

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f ( x ) = { x 2 + 3 x + 2 x + 2 , x 2 k , x = −2

k = −1

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f ( x ) = { e k x , 0 x < 4 x + 3 , 4 x 8

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f ( x ) = { k x , 0 x 3 x + 1 , 3 < x 10

k = 16 3

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In the following exercises, use the Intermediate Value Theorem (IVT).

Let h ( x ) = { 3 x 2 4 , x 2 5 + 4 x , x > 2 Over the interval [ 0 , 4 ] , there is no value of x such that h ( x ) = 10 , although h ( 0 ) < 10 and h ( 4 ) > 10 . Explain why this does not contradict the IVT.

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A particle moving along a line has at each time t a position function s ( t ) , which is continuous. Assume s ( 2 ) = 5 and s ( 5 ) = 2 . Another particle moves such that its position is given by h ( t ) = s ( t ) t . Explain why there must be a value c for 2 < c < 5 such that h ( c ) = 0 .

Since both s and y = t are continuous everywhere, then h ( t ) = s ( t ) t is continuous everywhere and, in particular, it is continuous over the closed interval [ 2 , 5 ] . Also, h ( 2 ) = 3 > 0 and h ( 5 ) = −3 < 0 . Therefore, by the IVT, there is a value x = c such that h ( c ) = 0 .

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[T] Use the statement “The cosine of t is equal to t cubed.”

  1. Write a mathematical equation of the statement.
  2. Prove that the equation in part a. has at least one real solution.
  3. Use a calculator to find an interval of length 0.01 that contains a solution.
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Apply the IVT to determine whether 2 x = x 3 has a solution in one of the intervals [ 1.25 , 1.375 ] or [ 1.375 , 1.5 ] . Briefly explain your response for each interval.

The function f ( x ) = 2 x x 3 is continuous over the interval [ 1.25 , 1.375 ] and has opposite signs at the endpoints.

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Consider the graph of the function y = f ( x ) shown in the following graph.

A diagram illustrating the intermediate value theorem. There is a generic continuous curved function shown over the interval [a,b]. The points fa. and fb. are marked, and dotted lines are drawn from a, b, fa., and fb. to the points (a, fa.) and (b, fb.).  A third point, c, is plotted between a and b. Since the function is continuous, there is a value for fc. along the curve, and a line is drawn from c to (c, fc.) and from (c, fc.) to fc., which is labeled as z on the y axis.
  1. Find all values for which the function is discontinuous.
  2. For each value in part a., state why the formal definition of continuity does not apply.
  3. Classify each discontinuity as either jump, removable, or infinite.
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Let f ( x ) = { 3 x , x > 1 x 3 , x < 1 .

  1. Sketch the graph of f .
  2. Is it possible to find a value k such that f ( 1 ) = k , which makes f ( x ) continuous for all real numbers? Briefly explain.

a.
A graph of the given piecewise function containing two segments. The first, x^3, exists for x < 1 and ends with an open circle at (1,1). The second, 3x, exists for x > 1. It beings with an open circle at (1,3).
b. It is not possible to redefine f ( 1 ) since the discontinuity is a jump discontinuity.

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Let f ( x ) = x 4 1 x 2 1 for x 1 , 1 .

  1. Sketch the graph of f .
  2. Is it possible to find values k 1 and k 2 such that f ( −1 ) = k and f ( 1 ) = k 2 , and that makes f ( x ) continuous for all real numbers? Briefly explain.
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Sketch the graph of the function y = f ( x ) with properties i. through vii.

  1. The domain of f is ( , + ) .
  2. f has an infinite discontinuity at x = −6 .
  3. f ( −6 ) = 3
  4. lim x −3 f ( x ) = lim x −3 + f ( x ) = 2
  5. f ( −3 ) = 3
  6. f is left continuous but not right continuous at x = 3 .
  7. lim x f ( x ) = and lim x + f ( x ) = +

Answers may vary; see the following example:
A graph of a piecewise function with several segments. The first is an increasing line that exists for x < -8. It ends at an open circle at (-8,-8). The second is an increasing curve that exists from -8 <= x < -6. It begins with a closed circle at (-8, 0 ) and goes to infinity as x goes to -6 from the left. The third is a closed circle at the point (-6, 3). The fourth is a line that exists from -6 < x <= 3. It begins with an open circle at (-6, 2) and ends with a closed circle at (3,2). The fifth is an increasing line starting with an open circle at (3,3). It exists for x > 3.

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Sketch the graph of the function y = f ( x ) with properties i. through iv.

  1. The domain of f is [ 0 , 5 ] .
  2. lim x 1 + f ( x ) and lim x 1 f ( x ) exist and are equal.
  3. f ( x ) is left continuous but not continuous at x = 2 , and right continuous but not continuous at x = 3 .
  4. f ( x ) has a removable discontinuity at x = 1 , a jump discontinuity at x = 2 , and the following limits hold: lim x 3 f ( x ) = and lim x 3 + f ( x ) = 2 .
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Practice Key Terms 9

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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