Question

, a function is defined and a closed interval is given. Decide whether the Mean Value Theorem applies to the given function on the given interval. If it does, find all possible values of \(c ;\) if not, state the reason. In each problem, sketch the graph of the given function on the given interval. $$ H(s)=s^{2}+3 s-1 ;[-3,1] $$

Short answer

The Mean Value Theorem applies; \( c = -1 \).

Step-by-step solution

Verify Conditions for Mean Value Theorem

To determine if the Mean Value Theorem (MVT) applies, we need to confirm two conditions for the function \( H(s) = s^2 + 3s - 1 \) on the interval \([-3, 1]\):1. **Continuity**: \( H(s) \) must be continuous on \([-3, 1]\). Since \( H(s) \) is a polynomial, it is continuous everywhere, therefore it is continuous on \([-3, 1]\).2. **Differentiability**: \( H(s) \) must be differentiable on \((-3, 1)\). Again, as a polynomial, \( H(s) \) is differentiable everywhere. Since both conditions are satisfied, the Mean Value Theorem applies.

Apply the Mean Value Theorem

According to the MVT, there exists at least one \( c \) in the open interval \((-3, 1)\) such that:\[H'(c) = \frac{H(1) - H(-3)}{1 - (-3)}\]

Calculate \( H(1) \) and \( H(-3) \)

First, calculate \( H(1) \):\[H(1) = (1)^2 + 3(1) - 1 = 1 + 3 - 1 = 3\]Next, calculate \( H(-3) \):\[H(-3) = (-3)^2 + 3(-3) - 1 = 9 - 9 - 1 = -1\]

Calculate the Derivative \( H'(s) \)

Find the derivative \( H'(s) \):\[H'(s) = 2s + 3\]

Solve for \( c \) Using MVT Equation

Substitute the values of \( H(1) \) and \( H(-3) \) into the MVT formula:\[H'(c) = \frac{3 - (-1)}{4} = \frac{4}{4} = 1\]Set \( H'(c) \) equal to 1:\[2c + 3 = 1\]Solve for \( c \):\[2c = 1 - 3 \c = \frac{-2}{2} \c = -1\]

Confirm \( c \) Lies Within the Interval \( (-3, 1) \)

The value \( c = -1 \) we found satisfies \(-3 < -1 < 1\), confirming it lies within the interval \((-3, 1)\). Therefore, it is a valid solution.

Sketch the Graph

Draw the graph of \( H(s) = s^2 + 3s - 1 \) over the interval \([-3, 1]\). It is a parabola opening upwards with a vertex below the \(x\)-axis. The secant line connecting \( (-3, H(-3)) \) and \( (1, H(1)) \) has a slope of 1, which equals the tangent slope at \( s = -1 \).

Key concepts

Polynomial Functions

In mathematics, a polynomial function is an expression that involves one or more variables raised to a power and combined using addition, subtraction, and multiplication. The function given in this exercise, \(H(s) = s^2 + 3s - 1\), is a polynomial function of degree 2, which is also known as a quadratic function. These types of functions are characterized by their parabolic shape.

Polynomials come in different degrees:

• Degree 0 polynomials, called constant functions (e.g., \(f(x) = 7\)).
• Degree 1 polynomials, called linear functions (e.g., \(f(x) = 3x + 2\)).
• Degree 2 polynomials, called quadratic functions (e.g., \(f(x) = x^2 - 4x + 4\)).

Understanding the degree of a polynomial helps to anticipate the shape of its graph: quadratic functions always graph to parabolas that open upwards or downwards.

For polynomial functions like \(H(s)\), graphing is a tool that provides insights into their behavior on different intervals and is particularly useful for understanding and applying concepts like the Mean Value Theorem.

Continuity and Differentiability

Continuity and differentiability are critical concepts in calculus, which ensure that we can use theories like the Mean Value Theorem effectively. For a function to be continuous, it must not have any breaks, jumps, or holes in its graph on the specified interval.

Polynomial functions, such as \(H(s) = s^2 + 3s - 1\), are continuous everywhere because they do not have breaks or jumps. This means on the interval \([-3, 1]\), \(H(s)\) is continuous and smooth. This makes it easier to analyze its behavior.

Differentiability refers to the ability to find a derivative at every point in a given interval. If a function is differentiable, it means it has a tangent line with a defined slope at each point in that interval. Again, polynomial functions are differentiable everywhere, meaning \(H(s)\) meets this condition on \((-3, 1)\).

Together, continuity and differentiability are vital conditions for utilizing the Mean Value Theorem, helping identify specific points where the slope of the tangent (instantaneous rate of change) equals the slope of the secant (average rate of change).

Graphing Functions

Graphing is an important aspect of understanding any function's behavior visually. In the exercise, we graph \(H(s) = s^2 + 3s - 1\) to gain insights into its properties over the interval \([-3, 1]\). The graph of this quadratic function is a parabola, which helps to visualize key properties such as the vertex, intercepts, and the direction it opens (upwards in this case).

Key points on a graph include:

• The intercepts, where the graph crosses the axes.
• The vertex, the highest or lowest point of a parabola.
• Symmetry axis, which for quadratics, divides the parabola into two mirror-image halves.

While graphing, it is important to mark critical points obtained from solving problems. For example, the point \(s = -1\) adds importance in confirming the application of the Mean Value Theorem, as it correlates with the slope of the secant line drawn from \((-3, H(-3))\) to \((1, H(1))\).