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A function \(y=f(x)\) and values of \(x_{0}\) and \(x_{1}\) are given. (a) Find the average rate of change of \(y\) with respect to \(x\) over the interval \(\left[x_{0}, x_{1}\right] .\) (b) Find the instantaneous rate of change of \(y\) with respect to \(x\) at the specified value of \(x_{0}\) (c) Find the instantaneous rate of change of \(y\) with respect to \(x\) a) The arbitrary value of \(x_{0} .\) (d) The average rate of change in part (a) is the slope of a certain secant line, and the instantaneous rate of change in part (b) is the slope of a certain tangent line. Sketch the graph of \(y=f(x)\) together with those two lines. $$ y=x^{3} ; x_{0}=1, x_{1}=2 $$

Short Answer

Expert verified
(a) 7, (b) 3; Secant slope is 7, tangent slope at \( x=1 \) is 3.

Step by step solution

01

Determine the formula for average rate of change

The average rate of change of a function \( y = f(x) \) over an interval \([x_0, x_1]\) is given by \( \frac{f(x_1) - f(x_0)}{x_1 - x_0} \). In this problem, our function is \( f(x) = x^3 \), with \( x_0 = 1 \) and \( x_1 = 2 \).
02

Calculate \( f(x_0) \) and \( f(x_1) \)

Substitute the values into the function:\[ f(1) = 1^3 = 1 \]\[ f(2) = 2^3 = 8 \]
03

Calculate the average rate of change

Using the results from Step 2, substitute into the formula for average rate of change:\[ \frac{f(2) - f(1)}{2 - 1} = \frac{8 - 1}{1} = 7 \]
04

Define instantaneous rate of change

The instantaneous rate of change at a point \( x = x_0 \) is the derivative of \( f(x) \) evaluated at \( x_0 \). This is equivalent to the slope of the tangent line at \( x_0 \).
05

Calculate the derivative of \( f(x) \)

The derivative of \( f(x) = x^3 \) is found using the power rule: \[ f'(x) = 3x^2 \]
06

Evaluate the derivative at \( x = x_0 \)

Substitute \( x_0 = 1 \) into the derivative:\[ f'(1) = 3(1)^2 = 3 \]
07

Interpret the results (Part a, b, and graph)

The average rate of change from \( x_0 \) to \( x_1 \) is 7, which is the slope of the secant line. The instantaneous rate of change at \( x = 1 \) is 3, which is the slope of the tangent line. A sketch of these lines on the graph of \( y = x^3 \) illustrates these slopes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Average Rate of Change
The average rate of change of a function over a certain interval provides a broad understanding of how the function's outputs change as the inputs progress from one point to another. Imagine tracing your finger along the curve of a hill. The average rate of change would represent the overall slope from the bottom to the top of that segment of the hill.
In mathematical terms, for a function \( y = f(x) \) over the interval \([x_0, x_1]\), the average rate of change is calculated as:
  • \( \frac{f(x_1) - f(x_0)}{x_1 - x_0} \)
Here, \( f(x_1) \) and \( f(x_0) \) indicate the function's value at the ending and starting points of the interval.

In our example with the function \( y = x^3 \), the interval from \( x_0 = 1 \) to \( x_1 = 2 \), the average rate of change is derived as \( 7 \). This means that, on average, for each unit increase in \( x \) from 1 to 2, \( y \) increases by 7 units.
Instantaneous Rate of Change
While the average rate offers a broad view over an interval, the instantaneous rate of change zooms in for a more granular look at a specific point. Think of it this way: if the average rate of change is like measuring your overall speed between two cities, the instantaneous rate is like checking your car's speedometer at a precise moment.
The instantaneous rate of change is synonymous with the derivative at a point. It tells us the slope of the tangent line to the curve at that very point. For any function \( y = f(x) \), to find the instantaneous rate of change at a point \( x_0 \), we compute the derivative and evaluate it at \( x_0 \).

In our case, given \( y = x^3 \), the derivative \( f'(x) = 3x^2 \) gives us the slope at any point \( x \). Evaluating it at \( x_0 = 1 \), we get \( f'(1) = 3 \). This value, 3, represents the instantaneous rate of change of the function at \( x = 1 \), indicating the slope of the tangent line at this point.
Derivative
The derivative of a function is a fundamental concept in calculus, crucial for understanding both average and instantaneous rates of change. In essence, it acts as a tool that allows us to dive deeper into the behavior of functions, capturing their rate of change, direction, and more.
Mathematically, the derivative of a function \( y = f(x) \) provides the slope of the curve at any point \( x \). It can be interpreted as a measure of how the function's output value changes in response to a small change in the input value. The power rule helps us find the derivative of polynomials like \( y = x^3 \):
  • \( f'(x) = 3x^2 \)
This derivative function \( f'(x) \) gives us the slope of the tangent line at any point on \( y = x^3 \).

As a practical illustration, if you substitute any \( x \) value into \( f'(x) \), you will know the function's rate of change at that point. In our scenario, when \( x = 1 \), the derivative \( f'(1) = 3 \) tells us that the function is rising steeply, which fits with our identified instantaneous rate of 3 at the same point.

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