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Show that the following continuous function has a root between 1 and 2. (round numerical answers to two decimal places.) g(x) = x2 − x − sin 8 x?

User Ganapat
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Final answer:

To prove there is a root for the function g(x) = x^2 - x - sin(8x) between 1 and 2, evaluate g(1) and g(2). Since g(1) < 0 and g(2) > 0 and g(x) is continuous, the Intermediate Value Theorem guarantees a root in the interval (1, 2).

Step-by-step explanation:

The student question pertains to proving that a continuous function has a root between two given points, specifically using the Intermediate Value Theorem (IVT). The function in question is g(x) = x2 − x − sin(8x). To show that there is a root between 1 and 2, we evaluate the function at these points:

  • g(1) = 12 - 1 - sin(8 * 1) = 1 - 1 - sin(8) ≈ 0 - 0 - 0.9893 ≈ -0.9893.
  • g(2) = 22 - 2 - sin(8 * 2) = 4 - 2 - sin(16) ≈ 2 - 0 ≈ 2.

Since g(1) is negative and g(2) is positive, and because the function g(x) is continuous, by the IVT, there must be a value c in the interval (1, 2) such that g(c) = 0, indicating the presence of a root.

User Sergey Sheleg
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