QUIZ: Intermediate Value Theorem A diabetic measures her blood sugar in the morning and it is 150. Later in the day she measures it again and it is 100. Can she use the IVT to prove that at some point her blood sugar was 120? '\\ yes 7' no 412345678910- Over what interval on the graph shown will the Intermediate Value Theorem (IVT) apply? (-5,5) [-5 ,0] or [0,5] f_\\. [_5 '5] ' " everywhere Over what interval on the graph shown will the Intermediate Value Theorem (IVT) apply? everywhere shown [ _5, 0] [5, .00001] (5, .00001) The IVT is often used to verify that a function has a zero. For the following graph what would be the proper way to state the IVT theorem to show that there is a zero in the range? -10 -9 -3 -7 -6 -5 -4 f(x) is continuous on [-5, 5] and let k be zero. Then there exists a number 0 such that, f(c) = 0. f(x) is continuous on [-6, -4] and let k be zero. Then there exists a number 0 such that, f(c) = 0. f(x) is continuous on [4, 5] and let k be zero. Then there exists a number 0 such that, f(c) = 0. f(x) is continuous on [-2, 5] and let k be zero. Then there exists a number c such that, f(c) = 0. Some mathematicians wanted to use the IVT as the definition of continuous. i.e. if the IVT does not hold then the function can not be continuous. For the following graph what is the counter example that shows that the IVT does not hold over a non-continuous graph? ' f(-4) = 1, f(2) = -1 so there must be a point where f(x) = .25. ' the IVT holds for the above graph. \\ f(-4) = 1, f(-2) = 1 so there must be a point where f(x) = .25. ' f(4) = -1, f(2) = -1 so there must be a point where f(x) = .25. Use the IVT to show thatf(x) = x3 has a zero in the interval [-1,1]. The IVT cannot be applied to find a zero. ' f(x) is continuous on [-1,1], f(-1)= -1, f(1) = 1, so we can not use the IVT to show there is a zero in [-1,1] ' f(x) is continuous on [1,2], f(1): 1, f(2) = 8, so we can use the IVT to show there is a zero, between [1,2]. \\ f(x) is continuous on [-1,1], f(-1)= -1, f(1) = 1, so we can use the IVT to show there is a zero in [-1,1] Use the NT to show that f(x)= |n(x+1) has a zero. You can use the NT to show that there is a zero in [-0.5, 2] You can not use the NT to show there is a zero in [0,2] The IVT cannot be applied to find a zero anywhere. ' You can use the NT to show there is a zero in [0,2] to show that f (x ) = sin (x) has a zero O The IVT cannot be applied to find a zero anywhere. O You can not use the IVT to show there is a zero in [-1,1] O You can use the IVT to show that there is a zero in [-1, 0] O You can use the IVT to show there is a zero in [-1,2]Use the NT to show thatf(x) = (x 2)3 + 1 has a zero. \\ The IVT cannot be applied to find a zero. f(x) is continuous on [-2,1], f(-2) 0, so we can use the IVT to show there is a zero in [-1,2] The IVT cannot be applied to find a zero. \\ f(x) is continuous on [0,1], f(O)