- Identify the interval \((-1, 0)\) as the location of the zero, since the function values \(f(-1)=-0.5\) and \(f(0)=2.5\) have opposite signs.
- Apply linear interpolation between the points \((-1, -0.5)\) and \((0, 2.5)\) to estimate the zero.
- The zero is found at \(x \approx -0.833\).
- Final Answer: \(\boxed{-\frac{5}{6}}\).
Explanation
- Identifying the Interval Using the Intermediate Value Theorem
The Intermediate Value Theorem (IVT) states that for a continuous function \(f(x)\), if \(f(a)\) and \(f(b)\) have opposite signs, then there must be at least one value \(c\) between \(a\) and \(b\) such that \(f(c) = 0\). We are given a table of values for a cubic function, which is continuous everywhere. Examining the values provided:
- \(f(-2) = -11.5\)
- \(f(-1) = -0.5\)
- \(f(0) = 2.5\)
We observe that \(f(-1) = -0.5\) (negative) and \(f(0) = 2.5\) (positive). Since the function changes sign between these input values, by the IVT, there must be a zero in the interval \((-1, 0)\).
- Approximating the Zero via Linear Interpolation
To approximate the zero of the function in the interval \((-1, 0)\), we can use linear interpolation between the two points \((-1, -0.5)\) and \((0, 2.5)\). We assume the function behaves linearly between these two points. The slope \(m\) of the line passing through \((-1, -0.5)\) and \((0, 2.5)\) is:
\[m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2.5 - (-0.5)}{0 - (-1)} = \frac{3}{1} = 3.\]
Using the point-slope form with the point \((0, 2.5)\), the equation of the line is:
\[y - 2.5 = 3(x - 0) \implies y = 3x + 2.5.\]
To find the approximate zero, set \(y = 0\):
\[0 = 3x + 2.5 \implies 3x = -2.5 \implies x = -\frac{2.5}{3} \approx -0.833.\]
Thus, the zero is approximately \(-0.833\).
- Conclusion
In summary, a zero can be found between the input values of \(-1\) and \(0\) because the function changes sign from \(-0.5\) to \(2.5\). The approximate zero of the function is \(-0.833\), or \(\boxed{-\frac{5}{6}}\).
Examples
Linear interpolation is a fundamental tool used in fields like engineering and data analysis to estimate values between known data points. For instance, if you have weather station data reporting temperatures at 8:00 AM and 9:00 AM, you can use linear interpolation to estimate the temperature at 8:30 AM. This technique assumes a smooth transition between observations, allowing you to make educated predictions in real-world scenarios where precise, continuous data might not be available at every instant.