Problem 37

Question

Suppose that $f(x)=x^{2}, x \in[-1,1]$. (a) Find the slope of the secant line connecting the points $(-1,1)$ and $(1,1)$. (b) Find a number $c \in(-1,1)$ such that $f^{\prime}(c)$ is equal to the slope of the secant line you computed in (a), and explain why such a number must exist in $(-1,1)$.

Step-by-Step Solution

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Answer

The slope of the secant line is 0, and the number is $c = 0$.

1Step 1: Identify the Points

The two points connecting the secant line are given by the function values at the endpoints of the interval. These points are $(-1, f(-1)) = (-1, 1)$ and $(1, f(1)) = (1, 1)$ as per the problem statement.

2Step 2: Calculate the Slope of the Secant Line

The slope of a secant line through two points $(x_1, y_1)$ and $(x_2, y_2)$ is calculated with $m = \frac{y_2 - y_1}{x_2 - x_1}$. For the points $(-1, 1)$ and $(1, 1)$, the slope is\[m = \frac{1 - 1}{1 - (-1)} = \frac{0}{2} = 0.\]

3Step 3: Find the Derivative of the Function

The derivative of a function $f(x) = x^2$ gives the slope of the tangent line at any point $x$. The derivative is calculated as\[f'(x) = \frac{d}{dx}(x^2) = 2x.\]

4Step 4: Find the Value of c

We need to find a $c$ such that $f'(c) = 0$. From $f'(x) = 2x$, setting $2x = 0$ gives the solution $x = 0.$ Thus, $c = 0$.

5Step 5: Apply the Mean Value Theorem

According to the Mean Value Theorem, if $f$ is continuous on $[-1, 1]$ and differentiable on $(-1, 1)$, then there is at least one point $c$ in $(-1, 1)$ such that $f'(c) = 0$. The function $f(x) = x^2$ meets these conditions, ensuring that some $c$ in the interval exists where $f'(c)$ matches the secant line slope.

Key Concepts

Understanding the Secant LineFunction Derivative BasicsCalculating Slopes Effectively

Understanding the Secant Line

A secant line is a straight line that cuts through a curve at two distinct points. To understand this concept better, think of it as a "shortcut" or "chord" that connects two points on a graph of a function. In our exercise, these points are

$(-1, f(-1)) = (-1, 1)$ and
$(1, f(1)) = (1, 1)$.

The slope of this secant line gives us an average rate of change of the function between those two points. It's calculated using the formula for slope: $ m = \frac{y_2 - y_1}{x_2 - x_1} $. In simpler terms, this formula tells us how much "upward shift" there is per "step across." In our exercise, the points

$y_2 = 1$ and $y_1 = 1$
$x_2 = 1$ and $x_1 = -1$,

provide a slope of $ m = 0 $. This result indicates that the secant line is horizontal, meaning no overall vertical change as we move from $-1$ to $1$.

Function Derivative Basics

The derivative of a function provides detailed information about the rate of change at any specific point along its curve. It's an important concept because while the secant line gives an average rate over an interval, the derivative tells us how steep the curve is at just one point. For the function $f(x) = x^2$, the derivative is calculated using calculus as $f'(x) = 2x$. This is derived by taking the limit of the average rates of change, effectively zeroing in on an infinitesimally small interval. Essentially, the derivative $2x$ explains how quickly the function's value is climbing or descending at any particular $x$ location. In our specific problem, we are looking for when this rate of change is zero. We find this using $f'(c) = 0$ which leads us to $c = 0$. This means that right at $x = 0$, our curve is perfectly flat, like a tangent line sitting right on the curve, which matches the slope of the secant line.

Calculating Slopes Effectively

When calculating slopes, clarity and precision are key. Whether it's for a secant or a tangent (derived from the derivative), knowing how to compute the slope accurately is essential for understanding the behavior of lines and curves on a graph. A slope $m$ is calculated as the "rise over run," using the expression $ m = \frac{y_2 - y_1}{x_2 - x_1} $. To put this into context with our exercise:

For the secant line through points $(-1, 1)$ and $(1, 1)$,\: this simplifies to $ m = 0 $.
For the derivative $f'(x) = 2x$, we see that the slope can vary based on the $x$ value.

Using the Mean Value Theorem, which applies since $f(x) = x^2$ is both continuous and differentiable, assures us there is a point $c$ in $-1 < c < 1$ where the derivative $f'(c)$ exactly matches this secant slope. Our calculation shows this point is when $ c = 0 $. Observing how the slope changes gives insight into the curve's direction and steepness, solidifying your understanding of the graph's shape.

Problem 37

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