Question

Max/min of area functions Suppose \(f\) is continuous on \([0, \infty)\) and \(A(x)\) is the net area of the region bounded by the graph of \(f\) and the \(t\) -axis on \([0, x] .\) Show that the local maxima and minima of \(A\) occur at the zeros of \(f .\) Verify this fact with the function \(f(x)=x^{2}-10 x\)

Short answer

Question: Show that the local maxima and minima of the net area function \(A(x)\) occur at the zeros of the function \(f(x) = x^2 - 10x\). Answer: To show that the local maxima and minima of the net area function \(A(x)\) occur at the zeros of the given function \(f(x) = x^2 - 10x\), we first found the derivative of \(A(x)\) using the Fundamental Theorem of Calculus, Part 1, which gave us \(A'(x) = f(x)\). Next, we found the critical points where \(A'(x) = f(x) = 0\), which led to the zeros of \(f(x)\) at \(x=0\) and \(x=10\). Using the Second Derivative Test, we evaluated \(A''(x)\) at the zeros and found that there is a local maximum at \(x=0\) and a local minimum at \(x=10\). Thus, the local maxima and minima of the net area function \(A(x)\) do occur at the zeros of \(f\).

Step-by-step solution

Derivative of \(A(x)\) (Fundamental Theorem of Calculus, Part 1)

Since \(A(x)\) is the net area of the region bounded by the graph of \(f\) and the \(t\)-axis on \([0, x]\), we can write \(A(x)\) as an integral: $$A(x)=\int_{0}^{x}{f(t)dt}$$ By the Fundamental Theorem of Calculus, Part 1, the derivative of \(A(x)\) with respect to \(x\) is given by \(f(x)\): $$A'(x)=f(x)$$

Find critical points of \(A(x)\)

To find the local maxima and minima of \(A(x)\), we need to find the critical points where \(A'(x) = 0\) or is undefined. In this case, \(A'(x) = f(x)\), and since \(f\) is continuous on \([0, \infty)\), it's always defined. Thus, we only need to find the zeros of \(f\): $$f(x)=0$$

Verify with the given function

We need to verify that local maxima and minima of \(A(x)\) occur at the zeros of \(f(x) = x^2 - 10x\). First, find the zeros of \(f(x)\): $$x^2 - 10x = x(x - 10) = 0 \Rightarrow x = 0, 10$$ The zeros of \(f(x)\) are \(x=0\) and \(x=10\). Next, let's find the second derivative of \(A(x)\) to use Second Derivative Test: $$A'(x) = f(x) = x^2 - 10x$$ $$A''(x) = 2x - 10$$ Now, we'll evaluate \(A''(x)\) at the zeros: 1. For \(x = 0\), \(A''(0) = 2(0) - 10 = -10\) 2. For \(x = 10\), \(A''(10) = 2(10) - 10 = 10\) Because \(A''(0) < 0\), we have a local maximum at \(x=0\). And since \(A''(10) > 0\), we have a local minimum at \(x=10\). So, the local maxima and minima of the net area function \(A(x)\) do indeed occur at the zeros of \(f\).

Key concepts

Maxima and Minima

Maxima and minima are all about finding the highest or lowest points on a graph. When we're dealing with a function like our area function \(A(x)\), local maxima and minima are the peaks and valleys where the graph changes direction. These occur where the derivative \(A'(x)\) is zero since that's where the slope of the tangent to the graph is flat.
This means we're finding out where the change in \(A(x)\) stops for a moment before changing the direction.

When using the Fundamental Theorem of Calculus, we learned that \(A'(x) = f(x)\). So, the critical points, which can be potential maxima or minima, happen when \(f(x) = 0\). In simple terms:

• Local maxima: Peaks on the graph where \(A(x)\) changes from increasing to decreasing.
• Local minima: Valleys on the graph where \(A(x)\) changes from decreasing to increasing.

Using the second derivative, \(A''(x)\), helps confirm if a critical point is a maximum or minimum. If \(A''(x) > 0\), it indicates a local minimum (a valley). If \(A''(x) < 0\), it indicates a local maximum (a peak). This step strengthens our understanding of the behavior at these critical points.

Critical Points

Critical points are where the action happens in calculus.
They are the \(x\)-values where the derivative of a function is zero or undefined. These points are special because they represent potential spots for local maxima or minima on the graph.

In our exercise, since \(A'(x) = f(x)\), the critical points are the zeros of \(f(x)\). For our function \(f(x) = x^2 - 10x\), solving \(f(x) = 0\) gives us the critical points \(x = 0\) and \(x = 10\).

• Critical points occur where the slope is zero, meaning the graph flattens out at these positions.
• They give us potential locations for maxima and minima, depending on the behavior of the function around these points.

Finding these critical points is crucial since they tell us where the rate of change in the function is zero. Once we find them, we can further analyze the function's behavior to determine the nature of these points, as we discussed with second derivatives.

Continuous Functions

A function being continuous is like having a smooth road without any holes or jumps. For the Fundamental Theorem of Calculus to work seamlessly, the function \(f\) must be continuous.
This means that there are no interruptions in the graph on the interval we're looking at.

A continuous function ensures that:

• We can draw the function without lifting our pencil from the paper.
• The integral, which represents area under the curve, exists and is well-defined.

In this problem, \(f(x) = x^2 - 10x\) is continuous on \([0, \infty)\), which means our calculations for \(A(x)\) and its derivatives are valid throughout the interval. This property allows us to apply different calculus rules reliably, like finding the derivative or solving integrals, without worrying about undefined points. That's why understanding continuous functions is vital for solving problems involving areas and derivatives effectively.