Working with area functions Consider the function ƒ and the points a, b, and c. (b) Graph ƒ and A. ƒ(𝓍) = 1/𝓍 ; a = 1 , b = 4 , c = 6
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First, understand that the function given is \( f(x) = \frac{1}{x} \), which is a hyperbola defined for \( x > 0 \).
Next, identify the points \( a = 1 \), \( b = 4 \), and \( c = 6 \) on the x-axis. These will be important for defining the intervals over which we consider the area function.
The area function \( A(x) \) is typically defined as the integral of \( f(t) \) from a fixed point \( a \) to a variable upper limit \( x \), so write \( A(x) = \int_{a}^{x} \frac{1}{t} \, dt \).
To graph \( f(x) \), plot the curve \( y = \frac{1}{x} \) for \( x \geq 1 \), noting that it decreases and approaches zero as \( x \) increases.
To graph the area function \( A(x) \), recognize that it represents the accumulated area under \( f(t) \) from \( t = 1 \) to \( t = x \). This area corresponds to the integral \( \int_{1}^{x} \frac{1}{t} \, dt \), which is related to the natural logarithm function.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Area Functions and the Definite Integral
An area function A(x) represents the accumulated area under the curve of a function ƒ from a fixed point a to a variable point x. It is defined using the definite integral, A(x) = ∫ₐˣ ƒ(t) dt, which measures the net area between the graph of ƒ and the x-axis over [a, x]. Understanding this concept is essential for interpreting and graphing area functions.
The function ƒ(x) = 1/x is a hyperbola with two branches, defined for x ≠ 0. It is positive and decreasing on the interval (0, ∞), approaching zero as x increases. Recognizing its shape and behavior helps in visualizing the area under the curve between points a, b, and c, which is crucial for sketching both ƒ and the area function A.
Relationship Between a Function and Its Area Function
The Fundamental Theorem of Calculus links a function ƒ and its area function A by stating that A'(x) = ƒ(x). This means the slope of the area function at any point x equals the value of ƒ at x. Understanding this relationship aids in graphing A by using the shape of ƒ and interpreting how the accumulated area changes over the interval.