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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 \(x\) at an 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, (c) \(3x^2\). Secant line slope is 7; tangent line slope is 3 at \(x=1\).

Step by step solution

01

Identify the Function and Interval

The given function is \( y = f(x) = x^3 \) and the interval is from \( x_0 = 1 \) to \( x_1 = 2 \).
02

Calculate Average Rate of Change

The average rate of change of \( y \) with respect to \( x \) over \([x_0, x_1]\) is given by the formula \( \frac{f(x_1) - f(x_0)}{x_1 - x_0} \). Plug in the values: \( f(2) = 2^3 = 8 \) and \( f(1) = 1^3 = 1 \). Thus, the average rate of change is \( \frac{8 - 1}{2 - 1} = 7 \).
03

Determine Instantaneous Rate of Change at \(x_0\)

The instantaneous rate of change at \( x_0 = 1 \) is found using the derivative of the function. The derivative of \( y = x^3 \) is \( f'(x) = 3x^2 \). Evaluating at \( x_0 = 1 \), we get \( f'(1) = 3 \cdot 1^2 = 3 \).
04

Determine Instantaneous Rate of Change at Arbitrary \(x\)

For any arbitrary \( x \), the instantaneous rate of change is given by the derivative \( f'(x) = 3x^2 \). So, for any \( x = a \), it is \( f'(a) = 3a^2 \).
05

Graph the Function and Lines

Plot the function \( y = x^3 \). Draw the secant line, which is a straight line passing through points \((1, 1)\) and \((2, 8)\) with a slope of 7. Then, draw the tangent line at \( x = 1 \), which is a line with slope 3 that just touches the curve at the point \((1, 1)\).

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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 provides a valuable snapshot of how a function behaves over a specific interval. For a function like \(y = f(x)\), it measures how much \(y\) changes as \(x\) moves from \(x_0\) to \(x_1\). This concept is similar to finding the slope of a straight line connecting two points on a graph.
If you imagine a hill's slope, the average rate of change tells you how steep the hill is between two markers. For our specific example, the average rate of change of \(y = x^3\) as \(x\) moves from \(1\) to \(2\) can be calculated using the formula:
  • \(\text{Average Rate of Change} = \frac{f(x_1) - f(x_0)}{x_1 - x_0} = \frac{8 - 1}{2 - 1} = 7\)
This tells us that for every unit increase in \(x\), the function \(y\) increases by 7 units over this interval. The line connecting the points \((1, 1)\) and \((2, 8)\) is known as a secant line, representing this average rate.
Instantaneous Rate of Change
The instantaneous rate of change is like a snapshot of the function's behavior at a specific point. It's the slope of the curve at an exact point, unlike the broader average rate over an interval. Imagine standing on a hill, the steepness at your feet is the instantaneous rate, whereas the average rate would describe from one point to another.
For our function \(y = x^3\), the derivative tells us the instantaneous rate of change. At \(x_0 = 1\), the rate is given by \
  • \(f'(x) = 3x^2\)
Evaluating this at \(x = 1\) gives us:
  • \(f'(1) = 3 \times 1^2 = 3\)
So, at \(x_0 = 1\), the rate of increase of \(y\) as \(x\) changes is 3. This is depicted by a tangent line to the curve at \((1, 1)\), illustrating the exact momentary slope there.
Derivatives
Derivatives are fundamental in calculus as they provide the tool to calculate the instantaneous rate of change. They are like a powerful magnifying glass for navigating the tiny changes within functions.
For any function, its derivative helps us understand how the function's output (\(y\)) will change with an infinitesimally small change in the input (\(x\)). In our example, the derivative of \(y = x^3\) is found as:
  • \(f'(x) = 3x^2\)
    • This indicates how \(y\) reacts to small shifts in \(x\). For an arbitrary \(x\), say \(x = a\), the derivative provides \(f'(a) = 3a^2\). Each point on the function can have its own instantaneous rate calculated this way, allowing us to model phenomena precisely, from population growth to physics problems.
      In drawing graphs, derivatives help create accurate tangent lines, which perfectly align with the curve at a given point, depicting real-time change.

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Most popular questions from this chapter

Newton's Law of Universal Gravitation states that the magnitude \(F\) of the force exerted by a point with mass \(M\) on a point with mass \(m\) is 2 $$ F=\frac{G m M}{r^{2}} $$ where \(G\) is a constant and \(r\) is the distance between the points. Assuming that the points are moving, find a formula for the instantancous rate of change of \(F\) with respect to \(r\)

Find the indicated derivative. $$\lambda=\left(\frac{a u+b}{c u+d}\right)^{6} ; \text { find } \frac{d \lambda}{d u} \quad(a, b, c, d \text { constants })$$

(a) Use a graphing utility to obtain the graph of the function \(f(x)=\sin x^{2} \cos x\) over the interval \([-\pi / 2, \pi / 2]\) (b) Use the graph in part (a) to make a rough sketch of the graph of \(f^{\prime}\) over the interval. (c) Find \(f^{\prime}(x)\), and then check your work in part (b) by using the graphing utility to obtain the graph of \(f^{\prime}\) over the interval. (d) Find the equation of the tangent line to the graph of \(f\) at \(x=1,\) and graph \(f\) and the tangent line together over the interval.

A function \(y=f(x)\) and an \(x\) -value \(x_{0}\) are given. (a) Find a formula for the slope of the tangent line to the graph of \(f\) at a general point \(x=x_{0}\). (b) Use the formula obtained in part (a) to find the slope of the tangent line for the given value of \(x_{0}\). $$f(x)=x^{2}+3 x+2 ; x_{0}=2$$

Find an equation for the tangent line to the graph at the specified value of \(x\) $$y=\sec ^{3}\left(\frac{\pi}{2}-x\right), x=-\frac{\pi}{2}$$

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