Ma 113 Functions And Inverse Functions, The Exponential Function And The Logarithm Worksheet Page 9
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41Worksheet # 7: The Intermediate Value Theorem
1. (a) State the Intermediate Value Theorem.
(b) Show that f (x) = x + x
1 has a zero in the interval [0, 1].
2. Let f (x) = e /(e
2). Show that f (0) < 1 < f (ln(4)).
Can you use the intermediate value theorem to conclude that there is a solution of f (x) = 1?
Can you find a solution to f (x) = 1?
3. Use the Intermediate Value Theorem to find an interval of length 1 in which a solution to the equation
2x + x = 5 must exist.
4. Show that there is some a with 0 < a < 2 such that a + cos(πa) = 4.
5. Show that the equation xe = 2 has a solution in the interval (0, 1).
Determine if the solution lies in the interval (0, 1/2) or (1/2, 1).
Continue to find an interval of length 1/8 which contains a solution of the equation xe = 2.
6. Find an interval for which the equation ln(x) = e
has a solution. Use the intermediate value theorem
to check your answer.
0 if x
0
7. Let f (x) =
be a piecewise function.
1 if x > 0
Although f ( 1) = 0 and f (1) = 1, f (x) = 1/2 for all x in its domain. Why doesn’t this contradict to
the Intermediate Value Theorem?
8. Prove that x =
1 has no solution.
9. Determine if the following are true or false.
(a) If f is continuous on [ 1, 1], f ( 1) =
2 and f (1) = 2, then f (0) = 0.
(b) If f is continuous on [ 1, 1], f ( 1) =
2 and the equation f (x) = 1 has no solution, then there
is no solution to f (x) = 0.
(c) If f is continuous on [ 1, 1], f ( 1) = 1, and the equation f (x) = 1 has no solution, then there is
no solution to f (x) = 2.
(d) If f is a function with domain [ 1, 1], f ( 1) =
2, f (1) = 2, and for each real number y, we can
find a solution to the equation f (x) = y, then f is continuous.
10. (Review) A particle moves along a line and its position after time t seconds is p(t) = 3t + 2t meters
to the right of the origin. Find the instantaneous velocity of the particle at t = 2.
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