3. (2 pts) Explain why the equation $4x^3 - 6x^2 + 3x - 2 = 0$ has a root between 1 and 2.
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Step 1: Let f(x) = 4x^3 - 6x^2 + 3x - 2. Show more…
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EXAMPLE 10: Show that there is a root of the equation 4x^3 - 6x^2 + 3x - 2 = 0 between 1 and 2. SOLUTION: Let f(x) = 4x^3 - 6x^2 + 3x - 2 = 0. We are looking for a solution of the given equation, that is, a number c between 1 and 2 such that f(c) = 0. Therefore, we take a = 1 and N = 2 in the Intermediate Value Theorem. We have: f(1) = 4 - 6 + 3 - 2 = 1 < 0 and f(2) = 32 - 24 + 6 - 2 = 12 > 0. Thus, f(1) < 0 < f(2); that is, N = 0 is a number between f(1) and f(2). Now, f is continuous since it is a polynomial, so the Intermediate Value Theorem says there is a number c between 1 and 2 such that f(c) = 0. In other words, the equation 4x^3 - 6x^2 + 3x - 2 = 0 has at least one root c in the open interval (1, 2). In fact, we can locate a root more precisely by using the Intermediate Value Theorem again. Since: f(1.2) = -0.128 < 0 and f(1.3) = 0.548 > 0 a root must lie between 1.2 and 1.3. A calculator gives, by trial and error: f(1.22) = -0.007008 < 0 and f(1.23) = 0.056068 > 0. So a root lies in the open interval (1.22, 1.23).
Adi S.
Prove that the equation x^5 - 3x - 1 = 0 has at least one root in the interval [1, 2].
Show that there is a root of the equation between 1 and 2. SOLUTION: Let f(x) = 3x^2 + 3x - 2. We are looking for a solution to the given equation, that is, a number c such that f(c) = 0. By the Intermediate Value Theorem, we have: f(1) = 3(1)^2 + 3(1) - 2 = 4 f(2) = 3(2)^2 + 3(2) - 2 = 16 Thus, f(1) < 0 < f(2); that means there is a number c between 1 and 2 such that f(c) = 0. Now, f(x) is continuous since it is a polynomial, and the Intermediate Value Theorem says that there is at least one root in the open interval (1, 2). In fact, we can locate the root more precisely by using f(1.6) = 4.272 and f(1.7) = 4.99. By the Intermediate Value Theorem again, there is a root between 1.6 and 1.7.
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