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

Answer

Question Analysis

OCR Text

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 )

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