Problem 51 - Thomas Calculus

Question

A continuous function \(y=f(x)\) is known to be negative at \(x=0\) and positive at \(x=1 .\) Why does the equation \(f(x)=0\) have at least one solution between \(x=0\) and \(x=1 ?\) Illustrate with a sketch.

Step-by-Step Solution

Answer

By the Intermediate Value Theorem, there is at least one solution for \(f(x)=0\) between \(x=0\) and \(x=1\).

Step 1: Identify the Information

The function \(y = f(x)\) is continuous, with \(f(0) < 0\) and \(f(1) > 0\). This means it starts below the x-axis at \(x=0\) and goes above the x-axis at \(x=1\).

Step 2: Apply the Intermediate Value Theorem

The Intermediate Value Theorem states that if \(f\) is continuous on the closed interval \([a,b]\) and \(f(a)\) and \(f(b)\) have opposite signs, then there exists at least one \(c\in(a,b)\) such that \(f(c)=0\).

Step 3: Verify Conditions for the Theorem

Check that the function is continuous on \([0,1]\), and that \(f(0) < 0\) and \(f(1) > 0\), which meet the conditions of the Intermediate Value Theorem.

Step 4: Conclusion

Since all conditions for the Intermediate Value Theorem are satisfied, we conclude that there is at least one \(c\) in the interval \((0,1)\) where \(f(c) = 0\), meaning \(f(x) = 0\) has at least one solution between \(x=0\) and \(x=1\).

Step 5: Illustrate with a Sketch

Draw the x-axis with points marked at \(x=0\) and \(x=1\). Sketch a continuous curve that starts below the x-axis at \(x=0\) and moves upwards to cross the x-axis, ending above the x-axis at \(x=1\). This illustrates visually that the curve must cross the x-axis at least once.

Key Concepts

Continuous Function

A continuous function is a type of function where small changes in the input result in small changes in the output. This can be visualized as a smooth curve with no breaks, jumps, or holes.
When considering a continuous function from a practical standpoint:

• The function can be drawn without lifting the pencil off the paper.
• Function values change gradually as you move along the x-axis.
• For the function to be considered continuous over an interval, it must be defined at every point within that interval and have no abrupt changes.

In our given problem with the function \(f(x)\), it was mentioned that \(f(x)\) is continuous on the interval \([0,1]\). This continuity ensures that the graph of \(f(x)\) smoothly connects the point \(x=0\) where it is negative, to the point \(x=1\) where it becomes positive.

Zero of a Function

The zero of a function, often referred to as the "root," is the value of \(x\) for which the function \(f(x)\) equals zero. When plotting the graph of a function, the zero corresponds to the point where the curve intersects the x-axis.Here’s why this is significant:

• Every zero of the function \(f(x)\) represents a solution to the equation \(f(x) = 0\).
• In the context of the Intermediate Value Theorem, if a continuous function changes from negative to positive values or vice versa over an interval, it must cross the x-axis at least once.
• Finding zeros can help solve real-world problems where you need to determine when a certain condition is met.

Applying this to our exercise, since \(f(0)\) is negative and \(f(1)\) is positive, the Intermediate Value Theorem guarantees there is at least one zero between \(x=0\) and \(x=1\). This provides assurance that a solution exists within the specified interval.

Graphical Representation

Graphical representation of functions helps us to visualize the behavior of functions and make sense of abstract mathematical concepts. In the context of this problem, drawing a graph is particularly useful to understand how the Intermediate Value Theorem works.A sketch constructed based on the problem would:

• Start at a point below the x-axis for \(x=0\) reflecting that \(f(0)<0\).
• End at a point above the x-axis for \(x=1\) reflecting that \(f(1)>0\).
• Show a continuous slope that rises subtly to cross the x-axis at least once between \(x=0\) and \(x=1\).

This visual demonstration connects the algebraic conditions provided by the Intermediate Value Theorem with the graphical nature of a continuous function crossing from negative to positive values. The act of crossing signifies the presence of at least one zero in the interval, which is exactly what we are trying to prove.