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Magazine Chapter 6: Problem 35
For the following exercises, refer to Table 8. $$\begin{array}{ccccccc}{x} & {1} & {2} & {3} & {4} & {5} & {6} \\ {f(x)} & {555} & {383} & {307} & {210} & {158} & {122}\end{array}$$ Use the intersect feature to find the value of \(x\) for which \(f(x)=250\) .
Short Answer
Expert verified
\( x \approx 3.588 \) for which \( f(x) = 250 \).
Step by step solution
01
Understanding the Table
The given table provides a set of values for the function \( f(x) \) at specific points \( x = 1, 2, 3, 4, 5, \) and \( 6 \). We need to determine which \( x \) value corresponds to \( f(x) = 250 \).
02
Identify Relevant Interval
Look at the values in the table to find where \( f(x) = 250 \) might lie. We see that \( f(3) = 307 \) and \( f(4) = 210 \). Since 250 is between 307 and 210, the corresponding \( x \) value should be between \( x = 3 \) and \( x = 4 \).
03
Estimate Value of \( x \)
Since the function values decrease as \( x \) increases in this interval and \( 250 \) is closer to \( 210 \) than to \( 307 \), \( x \) is likely closer to 4 than to 3. Based on estimation, \( x \) could be approximately 3.5 or a bit more.
04
Use Intersection Method
Consider \( f(x) \) as a linear function between \( x = 3 \) and \( x = 4 \). Use the intersection method (or linear interpolation) to find an exact \( x \) value such that \( f(x) = 250 \).
05
Linear Interpolation Calculation
Using linear interpolation: The formula is \( x = x_1 + \frac{(y - y_1)(x_2 - x_1)}{y_2 - y_1} \). Here, let \( x_1 = 3, x_2 = 4, y_1 = 307, y_2 = 210, y = 250 \).\[ x = 3 + \frac{(250 - 307)(4 - 3)}{210 - 307} = 3 + \frac{-57}{-97} \approx 3.588 \]
06
Conclusion
From the intersection method, the value of \( x \) for which \( f(x) = 250 \) is approximately \( x \approx 3.588 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Function Values
Function values are outputs of a function corresponding to specific inputs. In the context of our exercise, the table provides function values, noted as \( f(x) \), for various \( x \) values ranging from 1 to 6. For instance, when \( x = 1 \), \( f(x) = 555 \), indicating that 555 is the function value at that point. These values are part of a decreasing sequence, meaning each subsequent value is lower than the previous one.
- Function values provide information about how the function behaves at specific points.
- They help in identifying trends, such as whether the function is increasing or decreasing.
- They are essential for constructing graphs and performing accurate calculations, such as interpolation.
X-axis Intersection
The x-axis intersection, in the context of numerical tables, refers to finding the \( x \) value that aligns with a specified \( f(x) \) value. In the exercise, we aim to determine the \( x \) value when \( f(x) = 250 \). Since this exact value is not listed in the table, we must find where it would hypothetically 'cross' or intersect on the x-axis between two known points.
- This intersection is not literal, as points exist on a function, not physically intersecting on an axis.
- It involves understanding where this \( y \, \text{value} \) (250 in this case) would lie within the function's given \( x \) range.
- Calculating the x-axis intersection helps identify unknown values or verify potential solutions by insight into how a function progresses.
Linear Approximation
Linear approximation is a technique used to estimate the values of a function using line segments between points. In our case, we use linear interpolation as a type of approximation. This method assumes that the change between points is linear or, in simpler terms, progresses at a constant rate.
- Linear interpolation involves the assumption of linearity between two points to estimate unknown values.
- It is a valuable tool when function values are known only at specified points and not defined by an explicit formula.
- Using the formula \[ x = x_1 + \frac{(y - y_1)(x_2 - x_1)}{y_2 - y_1} \], we find the point where \( f(x) = 250 \).
Decreasing Functions
A decreasing function is where, as \( x \) increases, \( f(x) \) decreases, showing a downward trend. In our exercise, the function exhibits decreasing behavior, as evidenced by the declination of function values from 555 to 122 as \( x \) ascends from 1 to 6.
- Decreasing functions can be linear or nonlinear, but for segments considered linear between specific points.
- They inform us about the negative correlation between \( x \) and \( f(x) \), which can be vital for predicting unknown values.
- In this scenario, awareness of the decreasing trend aids in accurately narrowing down the range for \( f(x) = 250 \) and adjusting our calculations accordingly.
