Problem 1 - Calculus for Biology and Medicine

Question

Let $$ f(x)=x^{2}-1, \quad 0 \leq x \leq 2 $$ (a) Graph \(y=f(x)\) for \(0 \leq x \leq 2\). (b) Show that $$ f(0)<0

Step-by-Step Solution

Answer

Graph the function as a parabola between (0, -1) and (2, 3). Use the intermediate value theorem to show there's a \( c \) where \( f(c)=0 \) for \( 0

Step 1: Graph the Function

To graph the function \( y = f(x) = x^2 - 1 \) for \( 0 \leq x \leq 2 \), calculate and plot the points for several \( x \) values. This helps visualize the curve. For example, at \( x = 0 \), \( f(0) = 0^2 - 1 = -1 \), and at \( x = 2 \), \( f(2) = 2^2 - 1 = 3 \). The curve passes through these points, gradually increasing. This will form a parabola that opens upwards.

Step 2: Evaluate Function at Boundary Points

Check the function values at the boundaries: Calculate \( f(0) = 0^2 - 1 = -1 \) and \( f(2) = 2^2 - 1 = 3 \). These calculations confirm that \( f(0) < 0 \) and \( f(2) > 0 \).

Step 3: Apply the Intermediate Value Theorem

According to the intermediate value theorem, if a continuous function \( f(x) \) on a closed interval \([a, b]\) results in \( f(a) \) and \( f(b) \) having opposite signs, then there must be at least one \( c \) in \( (a, b) \) where \( f(c) = 0 \). Given that \( f(x) = x^2 - 1 \) is continuous, and \( f(0) < 0 \) while \( f(2) > 0 \), there exists some \( c \in (0, 2) \) such that \( f(c) = 0 \).

Key Concepts

Continuous Functions

A continuous function is one where small changes in the input result in small changes in the output. This means there are no jumps, breaks, or holes in the graph of the function. Understanding continuous functions is crucial for applying the Intermediate Value Theorem (IVT).For a function to be continuous on a given interval, it must meet three conditions:

• The function is defined at every point in the interval.
• The limit of the function as it approaches any point within the interval is equal to the function's value at that point.
• There are no discontinuities, which mean no sudden jumps or gaps in the values.

The function given in the problem, \( f(x) = x^2 - 1 \), is continuous for all real numbers since it's a polynomial. Polynomial functions are smooth and unbroken, making them ideal candidates for using the IVT. In our exercise, investigating the function on the interval \( \[0, 2\] \), we can confidently assert that \( f(x) \) does not exhibit any interruptions, as it is squared and shifted vertically by 1 to ensure a smooth progression.

Graphing Quadratic Functions

Graphing quadratic functions is a foundational skill in mathematics, allowing students to visualize the behavior of polynomials of degree two. A quadratic function typically takes the form \( y = ax^2 + bx + c \). Here, our specific function is \( y = x^2 - 1 \), where \( a = 1, b = 0, \text{ and } c = -1 \).To begin graphing, consider the basic shape: a parabola. This parabola opens upwards since \( a = 1 \). It curves smoothly from \( (-1) \) upwards through the points calculated for the specific interval.

Key Points to Plot

• The vertex: This is the turning point. For our function, it's at \((0, -1)\) since we start at \( x = 0 \).
• Evaluate points within the interval, such as \( f(1) = 1^2 - 1 = 0 \), to get more accuracy.
• Boundaries: Use boundary points like \( x = 0 \) and \( x = 2 \), where \( f(0) = -1 \) and \( f(2) = 3 \).

These steps help build a complete picture of how the function behaves across its defined range, showcasing the gradual rise from negative to positive values as \( x \) increases from 0 to 2.

Boundary Values

Boundary values in mathematical problems are the endpoints of a given interval. They're significant when evaluating functions to determine behavior at the limits.For \( f(x) = x^2 - 1 \) on the interval \( [0,2] \), boundary values are \( x = 0 \) and \( x = 2 \). Check these by:

• Calculating \( f(0) \) which gives \(-1\).
• Calculating \( f(2) \) which gives \(3\).

These boundary values are pivotal for applying the Intermediate Value Theorem. Since \( f(0) < 0 \) and \( f(2) > 0 \), the IVT tells us there must be some \( c \) in \((0,2)\) where \( f(c) = 0 \). Thus, knowing these boundary values allows for effectively confirming the presence of a root in this interval without detailed calculations of every value in between. They show the transition from negative to positive, confirming that zero must be crossed somewhere in the middle.

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Key Concepts

Continuous Functions

A continuous function is one where small changes in the input result in small changes in the output. This means there are no jumps, breaks, or holes in the graph of the function. Understanding continuous functions is crucial for applying the Intermediate Value Theorem (IVT).For a function to be continuous on a given interval, it must meet three conditions:

• The function is defined at every point in the interval.
• The limit of the function as it approaches any point within the interval is equal to the function's value at that point.
• There are no discontinuities, which mean no sudden jumps or gaps in the values.

The function given in the problem, \( f(x) = x^2 - 1 \), is continuous for all real numbers since it's a polynomial. Polynomial functions are smooth and unbroken, making them ideal candidates for using the IVT. In our exercise, investigating the function on the interval \( \[0, 2\] \), we can confidently assert that \( f(x) \) does not exhibit any interruptions, as it is squared and shifted vertically by 1 to ensure a smooth progression.

Graphing Quadratic Functions

Graphing quadratic functions is a foundational skill in mathematics, allowing students to visualize the behavior of polynomials of degree two. A quadratic function typically takes the form \( y = ax^2 + bx + c \). Here, our specific function is \( y = x^2 - 1 \), where \( a = 1, b = 0, \text{ and } c = -1 \).To begin graphing, consider the basic shape: a parabola. This parabola opens upwards since \( a = 1 \). It curves smoothly from \( (-1) \) upwards through the points calculated for the specific interval.

Key Points to Plot

• The vertex: This is the turning point. For our function, it's at \((0, -1)\) since we start at \( x = 0 \).
• Evaluate points within the interval, such as \( f(1) = 1^2 - 1 = 0 \), to get more accuracy.
• Boundaries: Use boundary points like \( x = 0 \) and \( x = 2 \), where \( f(0) = -1 \) and \( f(2) = 3 \).

These steps help build a complete picture of how the function behaves across its defined range, showcasing the gradual rise from negative to positive values as \( x \) increases from 0 to 2.

Boundary Values

Boundary values in mathematical problems are the endpoints of a given interval. They're significant when evaluating functions to determine behavior at the limits.For \( f(x) = x^2 - 1 \) on the interval \( [0,2] \), boundary values are \( x = 0 \) and \( x = 2 \). Check these by:

• Calculating \( f(0) \) which gives \(-1\).
• Calculating \( f(2) \) which gives \(3\).

These boundary values are pivotal for applying the Intermediate Value Theorem. Since \( f(0) < 0 \) and \( f(2) > 0 \), the IVT tells us there must be some \( c \) in \((0,2)\) where \( f(c) = 0 \). Thus, knowing these boundary values allows for effectively confirming the presence of a root in this interval without detailed calculations of every value in between. They show the transition from negative to positive, confirming that zero must be crossed somewhere in the middle.