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91Ó°ÊÓ

Using the intermediate value theorem, determine, if possible, whether the function \(f\) has a real zero between a and \(b\). $$f(x)=3 x^{2}-2 x-11 ; a=2, b=3$$

Short Answer

Expert verified
Yes, \(f(x)\) has a real zero between \(2\) and \(3\).

Step by step solution

01

- Evaluate the function at the endpoints

First, find the values of the function at the endpoints given: Evaluate at \(a = 2\): \[f(2) = 3(2)^2 - 2(2) - 11 = 3(4) - 4 - 11 = 12 - 4 - 11 = -3\] Then evaluate at \(b = 3\):\[f(3) = 3(3)^2 - 2(3) - 11 = 3(9) - 6 - 11 = 27 - 6 - 11 = 10\]
02

- Analyze the sign of the function values

Since \(f(2) = -3\) and \(f(3) = 10\), the values of \(f(x)\) at the endpoints are of opposite signs (\( -3 < 0\) and \(10 > 0\)). This indicates a change in sign.
03

- Apply the Intermediate Value Theorem

The Intermediate Value Theorem states that if \(f(x)\) is continuous on the interval \([a, b]\) and \(f(a)\) and \(f(b)\) have opposite signs, then \(f(x)\) has at least one real zero between \(a\) and \(b\). Since \(f(x) = 3x^2 - 2x - 11\) is a polynomial, it is continuous everywhere. Therefore, by the Intermediate Value Theorem, \(f(x)\) has at least one real zero between \(2\) and \(3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polynomial Functions
Polynomial functions are mathematical expressions consisting of variables and coefficients. A typical polynomial function looks something like this: \[ f(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + \ldots + a_{1}x + a_{0} \] Each term has a variable (\(x\)) raised to a non-negative integer power () and each variable is multiplied by a coefficient (). Polynomial functions can be simple like \(f(x) = 2x + 1\) or complex like the one in our exercise: \(f(x) = 3x^2 - 2x - 11\). Polynomials are powerful because they’re continuous and smooth, meaning you can graph them without lifting your pen. Evaluating polynomial functions at given points can be easily done by substituting the point (e.g., \(x = 2\)) into the function. This is exactly what we did in the exercise with \(f(2)\) and \(f(3)\).
Real Zeros
Real zeros of a function are the points where the function intersects the x-axis. In other words, they are the solutions where \[ f(x) = 0 \] To find real zeros, you can either solve the equation algebraically or use tools like graphing and the Intermediate Value Theorem (IVT). In the given problem, the Intermediate Value Theorem is useful. The IVT helps us determine if there's a point where the function's value changes from positive to negative (or vice versa) without actually solving the equation analytically. For example, if \(f(2) = -3\) and \(f(3) = 10\), we notice a sign change between 2 and 3, indicating there must be at least one real zero in this interval, provided the function is continuous.
Continuity
Continuity in functions means that the function's graph has no breaks, holes, or gaps. A function like our polynomial \(f(x) = 3x^2 - 2x - 11\) is continuous because it’s a polynomial, and all polynomials are continuous on the entire real number line. This trait is crucial when using the Intermediate Value Theorem. For IVT to apply, the function must be continuous on the interval we are inspecting. If it is, and there is a sign change within that interval, we can confidently state there is a real zero within. When you’re dealing with any function, always check for continuity before applying IVT. For polynomials, this is straightforward, but for other types of functions, examining the domain and potential points of discontinuity is essential.

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