Problem 3 - Calculus

Question

A function is defined and a closed in terval 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. $$ f(x)=x^{2}+x ;[-2,2] $$

Step-by-Step Solution

Answer

MVT applies and \(c = 0\) is the value where \(f'(c) = 1\).

Step 1: Verify Continuity

For the Mean Value Theorem (MVT) to apply, the function must be continuous on the closed interval \([-2, 2]\). The function given is \(f(x) = x^2 + x\), a polynomial, which is continuous everywhere. Thus, it is continuous on \([-2, 2]\).

Step 2: Verify Differentiability

MVT also requires the function to be differentiable on the open interval \((-2, 2)\). Again, since \(f(x) = x^2 + x\) is a polynomial function, it is differentiable everywhere, specifically on \((-2, 2)\).

Step 3: Apply the Mean Value Theorem

Because \(f(x)\) meets both criteria of continuity on \([-2, 2]\) and differentiability on \((-2, 2)\), we can apply the Mean Value Theorem. According to MVT, there exists at least one \(c\) in \((-2, 2)\) such that \(f'(c) = \frac{f(2) - f(-2)}{2 - (-2)}\).

Step 4: Calculate \(f(a)\) and \(f(b)\)

First, calculate \(f(-2) = (-2)^2 + (-2) = 4 - 2 = 2\) and \(f(2) = 2^2 + 2 = 4 + 2 = 6\).

Step 5: Use the MVT Formula

Substitute back into the MVT formula: \(f'(c) = \frac{6 - 2}{2 - (-2)} = \frac{4}{4} = 1\).

Step 6: Find \(f'(x)\) and Solve for \(c\)

First, find the derivative of \(f(x)\): \(f'(x) = 2x + 1\). Set this equal to 1: \(2x + 1 = 1\). Solve for \(x\): \(2x = 0\) giving \(x = 0\).\Since \(0 \in (-2, 2)\), \(c = 0\).

Step 7: Sketch the Graph

The graph of \(f(x) = x^2 + x\) is a parabola opening upwards with its vertex at \(-\frac{b}{2a} = -\frac{1}{2}\). Sketch this graph focusing on the interval \([-2, 2]\), showing \(f(-2) = 2\) and \(f(2) = 6\), with a tangent line slope of 1 at \(c = 0\).

Key Concepts

Continuity of Functions

Continuity of a function is a crucial concept in calculus, helping us understand the behavior of functions without interruptions. A function is said to be continuous over an interval if, roughly speaking, you can draw its graph without lifting your pencil from the paper.
For the Mean Value Theorem (MVT) to be applied, the function must be continuous over a closed interval \([-2, 2]\).
The function given in the exercise was \(f(x) = x^2 + x\), which is a polynomial function. **Why Polynomials Are Continuous:**

• Polynomials are one of the simplest types of functions that are always continuous over the entire set of real numbers.
• No breaks or gaps occur in their graphs.

In our exercise, since \(f(x)\) is a polynomial, it is continuous over the interval \([-2, 2]\).
This means we have confirmed the first essential criterion for applying the MVT.

Differentiability of Functions

Differentiability is another important concept which tells us if a function has a derivative; this derivative represents the function's instantaneous rate of change at any point in an interval. For the Mean Value Theorem to be applied, a function must be differentiable on an open interval \((-2, 2)\). A differentiable function means you can calculate its slope at any point in the given interval.
**Differentiability ensures:**

• There are no sharp points or corners on the graph in the open interval.
• The slope can be defined every moment in the interval.

Returning to our polynomial function, \(f(x) = x^2 + x\), we confirm that it is differentiable everywhere:

• Polynomial functions are smooth and have well-defined derivatives everywhere.

Thus, \(f(x) = x^2 + x\) is differentiable on the interval \((-2, 2)\), satisfying the second condition to use the MVT.

Polynomial Functions

Polynomial functions are essential in calculus due to their simplicity and ease of manipulation. They are functions composed of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. The general form of a polynomial is:
\[f(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0\]
**Characteristics of Polynomial Functions:**

• They are continuous and differentiable everywhere.
• Their graphs are smooth and extend infinitely in both directions unless otherwise restricted.

Analyzing the given polynomial \(f(x) = x^2 + x\), we find:

• It is a quadratic polynomial, a simple form of polynomial with degree 2.
• These functions form a parabola, a symmetric curve opening either upward or downward.

In our problem, understanding the nature of polynomial functions helped to establish two fundamental facts: the given function \(f(x)\) is both continuous and differentiable over the specified intervals, fulfilling the criteria necessary to apply the Mean Value Theorem. This ensures you can find at least one value \(c\) within the interval where the instantaneous rate of change (derivative) equals the average rate of change over \([-2, 2]\).