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2.4, #6, Given Problem.
| Sketch the graph of a function that has a jump discontinuity at x = 2 and a removable discontinuity at x = 4, but is continuous elsewhere. | 2.4, #6, Given Problem. |
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Here we have a graph which has the desired discontinuities.
Please note: There are an infinite number of graphs which could satisfy this set of requirements. |
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2.4, #26, Given Problem. |
 is continuous for
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We know that the sine function is continuous everywhere. |
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Because the function is everywhere continuous, the direct substitution method can be used. |
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We can now make this key substitution, and finish the problem. |
Given that  .
(a) Prove that the equation has at least one real root. (b) Use your calculator to find an interval of length 0.01 that contains a root. |
2.4, #40. Given Problem. |
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Part (a).
By Theorem 5 (p. 122) we know that polynomials are continuous everywhere. We know that at -1 the function is negative, and at 0 the function is positive. Add this to the fact that the function is continuous on this interval and the inevitable conclusion is that there must be a zero or root in the interval [0,1] |
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A partial table of values shows that the root must be between -0.87 and -0.88. |
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