Question

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kuta software - infinite calculus
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average rates of change
date____________ period__
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 )

Explanation:

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

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

Using \(\frac{f(3) - f(0)}{3 - 0}\), substitute the values:
\(\frac{7 - 1}{3 - 0} = \frac{6}{3} = 2\).

Step 1: Recall the average rate of change formula

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

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

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

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

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

Step 4: Apply the average rate of change formula

Using \(\frac{f(-2) - f(-3)}{-2 - (-3)}\), substitute the values:
First, simplify the denominator: \( -2 - (-3) = -2 + 3 = 1 \).
Then, the numerator: \( \frac{1}{4} - \frac{1}{5} = \frac{5 - 4}{20} = \frac{1}{20} \).
Thus, \(\frac{\frac{1}{20}}{1} = \frac{1}{20}\).

Answer:

2

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