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kuta software - infinite calculus average rates of change for each prob…

\( 2 \) ### Problem 2: \( v = -\frac{1}{x - 2} \); \([-3, -2]\)

カテゴリ: calculus 更新日: 2026-02-06

問題

kuta software - infinite calculus
average rates of change
for each problem, find the average rate of change of the function over the given interval.

  1. ( y = x^2 - x + 1 ); ( 0, 3 )
  2. ( v = -dfrac{1}{x - 2} ); ( -3, -2 )

解法ステップ

  1. Understand the question

    kuta software - infinite calculus
    average rates of change
    for each problem, find the average rate of change of the function over the given interval.

    1. ( y = x^2 - x + 1 ); ( 0, 3 )
    2. ( v = -dfrac{1}{x - 2} ); ( -3, -2 )
  2. Response
    Problem 1: \( y = x^2 - x + 1 \); \([0, 3]\)
  3. Explanation

    Step 1: Recall the average rate of change formula

    The average rate of change of a function \( y = f(x) \) over the interval \([a, b]\) is given by \(\frac{f(b) - f(a)}{b - a}\). Here, \( a = 0 \), \( b = 3 \), and \( f(x) = x^2 - x + 1 \).

    Step 2: Calculate \( f(0) \)

    Substitute \( x = 0 \) into \( f(x) \):
    \( f(0) = (0)^2 - 0 + 1 = 1 \).

    Step 3: Calculate \( f(3) \)

    Substitute \( x = 3 \) into \( f(x) \):
    \( f(3) = (3)^2 - 3 + 1 = 9 - 3 + 1 = 7 \).

    Step 4: Apply the average rate of change formula

    \(\frac{f(3) - f(0)}{3 - 0} = \frac{7 - 1}{3} = \frac{6}{3} = 2\).

  4. Explanation

    Step 1: Recall the average rate of change formula

    The average rate of change of a function \( v = g(x) \) over \([a, b]\) is \(\frac{g(b) - g(a)}{b - a}\). Here, \( a = -3 \), \( b = -2 \), and \( g(x) = -\frac{1}{x - 2} \).

    Step 2: Calculate \( g(-3) \)

    Substitute \( x = -3 \) into \( g(x) \):
    \( g(-3) = -\frac{1}{-3 - 2} = -\frac{1}{-5} = \frac{1}{5} \).

    Step 3: Calculate \( g(-2) \)

    Substitute \( x = -2 \) into \( g(x) \):
    \( g(-2) = -\frac{1}{-2 - 2} = -\frac{1}{-4} = \frac{1}{4} \).

    Step 4: Apply the average rate of change formula

    \(\frac{g(-2) - g(-3)}{-2 - (-3)} = \frac{\frac{1}{4} - \frac{1}{5}}{-2 + 3}\).
    First, simplify the numerator: \(\frac{1}{4} - \frac{1}{5} = \frac{5 - 4}{20} = \frac{1}{20}\).
    The denominator is \( 1 \). Thus, \(\frac{\frac{1}{20}}{1} = \frac{1}{20}\).

  5. Final answer

    \( 2 \)

    Problem 2: \( v = -\frac{1}{x - 2} \); \([-3, -2]\)

答え

Response

Problem 1: \( y = x^2 - x + 1 \); \([0, 3]\)

Explanation

Step 1: Recall the average rate of change formula

The average rate of change of a function \( y = f(x) \) over the interval \([a, b]\) is given by \(\frac{f(b) - f(a)}{b - a}\). Here, \( a = 0 \), \( b = 3 \), and \( f(x) = x^2 - x + 1 \).

Step 2: Calculate \( f(0) \)

Substitute \( x = 0 \) into \( f(x) \):
\( f(0) = (0)^2 - 0 + 1 = 1 \).

Step 3: Calculate \( f(3) \)

Substitute \( x = 3 \) into \( f(x) \):
\( f(3) = (3)^2 - 3 + 1 = 9 - 3 + 1 = 7 \).

Step 4: Apply the average rate of change formula

\(\frac{f(3) - f(0)}{3 - 0} = \frac{7 - 1}{3} = \frac{6}{3} = 2\).

Answer

\( 2 \)

Problem 2: \( v = -\frac{1}{x - 2} \); \([-3, -2]\)

Explanation

Step 1: Recall the average rate of change formula

The average rate of change of a function \( v = g(x) \) over \([a, b]\) is \(\frac{g(b) - g(a)}{b - a}\). Here, \( a = -3 \), \( b = -2 \), and \( g(x) = -\frac{1}{x - 2} \).

Step 2: Calculate \( g(-3) \)

Substitute \( x = -3 \) into \( g(x) \):
\( g(-3) = -\frac{1}{-3 - 2} = -\frac{1}{-5} = \frac{1}{5} \).

Step 3: Calculate \( g(-2) \)

Substitute \( x = -2 \) into \( g(x) \):
\( g(-2) = -\frac{1}{-2 - 2} = -\frac{1}{-4} = \frac{1}{4} \).

Step 4: Apply the average rate of change formula

\(\frac{g(-2) - g(-3)}{-2 - (-3)} = \frac{\frac{1}{4} - \frac{1}{5}}{-2 + 3}\).
First, simplify the numerator: \(\frac{1}{4} - \frac{1}{5} = \frac{5 - 4}{20} = \frac{1}{20}\).
The denominator is \( 1 \). Thus, \(\frac{\frac{1}{20}}{1} = \frac{1}{20}\).

Answer

\( \frac{1}{20} \)

Question Image

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Question Analysis

Subject mathematics
Sub Subject calculus
Education Level high school
Difficulty unspecified
Question Type with image, calculation, multiple (since there are two numbered problems: 1 and 2)
Multi Question No
Question Count 1
Analysis Status completed
Analyzed At 2026-02-06T20:11:50

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kuta software - infinite calculus
average rates of change
for each problem, find the average rate of change of the function over the given interval.
1) ( y = x^2 - x + 1 ); ( 0, 3 )
2) ( v = -dfrac{1}{x - 2} ); ( -3, -2 )

関連トピック

mathematicscalculuswith image, calculation, multiple (since there are two numbered problems: 1 and 2)high schoolliked-answerturns-1

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