Question

The linear function \(f(x)=3-x\) is decreasing on the interval \([0,3] .\) Is its area function for \(f\) (with left endpoint 0 ) increasing or decreasing on the interval [0,3]\(?\) Draw a picture and explain.

Short answer

Answer: The area function for the linear function f(x) = 3-x with the left endpoint at 0 is decreasing on the interval [0, 3].

Step-by-step solution

Find the area function of the given linear function

The area function is the definite integral of \(f(x)\) with the left endpoint at 0 and the variable endpoint at x. So, we will find: $$A(x) = \int_0^x f(t)dt = \int_0^x (3-t)dt$$

Calculate the definite integral

Now, we will calculate the definite integral to find the area function: $$A(x) = \int_0^x (3-t)dt = \left[ 3t - \frac{1}{2}t^2 \right]_0^x = 3x - \frac{1}{2}x^2$$

Analyze the slope of the area function on the interval [0, 3]

To determine if the area function is increasing or decreasing over the interval \([0, 3]\), we analyze the slope of the area function on this interval. To do this, we will find the derivative of the area function: $$A'(x) = \frac{d}{dx}(3x - \frac{1}{2}x^2) = 3 - x$$ Now, we will analyze the derivative of the area function over the interval [0, 3]. On this interval, \(A'(x)\) is a linear function with a negative slope of -1, meaning that it will be decreasing over the interval [0, 3].

Draw a picture and explain

To visualize the area function and its derivative, we can draw two separate graphs: one for \(f(x)\) and one for the area function \(A(x)\). The graph of the linear function \(f(x) = 3 - x\) will show a straight line with a negative slope. The graph of the area function \(A(x) = 3x - \frac{1}{2}x^2\) will show a parabolic function that opens downward. As we can see from the graph of the area function, its slope decreases as we go from left to right on the interval [0, 3]. This supports our conclusion that the area function is decreasing on this interval.

Conclusion

In conclusion, the area function for the linear function \(f(x) = 3-x\) with the left endpoint at 0 is decreasing on the interval [0, 3]. The graph of the area function that we drew supports this conclusion.

Key concepts

Definite Integral

A definite integral is a mathematical concept that calculates the accumulated area underneath a curve within a specified interval. In simpler terms, it measures the total region that a function covers between two points on its graph. For example, when we work with the function \( f(x) = 3-x \) from 0 to 3, the definite integral helps us find the accumulated area under this line, which can be understood as the area below the curve of the function and above the x-axis on the interval

• This involves evaluating the integral \( \int_0^3 (3-t) \, dt \).
• Such tasks help in understanding the behavior and properties of functions over specific intervals.

Understanding definite integrals is crucial in solving complex calculus problems, especially when determining total quantities in physical applications, such as displacement and work.

Linear Function

Linear functions are the simplest forms of polynomial functions. They have the standard form \( f(t) = mt + b \), where \( m \) is the slope and \( b \) is the y-intercept. In our case, \( f(t) = 3-t \) is a linear function, which represents a straight line with a slope of -1 and a y-intercept of 3.

• The negative slope means as \( t \) increases, \( f(t) \) decreases.
• This is why \( f(t) = 3-t \) is decreasing across its entire domain.

Despite their simplicity, linear functions are foundational in calculus and are often the starting point for learning about more complex functions.

Area Function

Area functions are constructed using definite integrals to provide a new function that describes the area under a curve from a fixed starting point to any other point. Essentially, it tells us the 'total area covered' up to a point. In our provided problem, the area function \( A(x) \) is derived from \( f(x) = 3-x \) as:

• The integral \( A(x) = \int_0^x (3-t) \, dt \) results in \( A(x) = 3x - \frac{1}{2}x^2 \).
• This provides a parabolic function that helps analyze changes in area as x varies from 0 to 3.

Understanding how the area function behaves, such as whether it increases or decreases over an interval, provides valuable insights into the nature of the original function it's based on.

Derivative

The derivative of a function represents the rate of change or slope at any given point of the function. Simply put, it tells us how the output of a function changes as the input changes. In our exercise, the derivative of the area function \( A(x) = 3x - \frac{1}{2}x^2 \) is crucial. It is calculated as:

• \( A'(x) = \frac{d}{dx}(3x - \frac{1}{2}x^2) = 3 - x \).
• This derivative shows us the slope of the area function at any point \( x \).
• If \( A'(x) \) is positive, the area function is increasing; if it's negative, it's decreasing.

In the interval [0,3], \( 3-x \) demonstrates a decreasing slope, implying the area function is decreasing throughout this interval. This explains the overall behavior of cumulative areas on that segment.