kuta software - infinite calculus name_____________________________ ave…
2 ### Problem 2: \( v = -\frac{1}{x - 2} \); \([-3, -2]\)
2 ### Problem 2: \( v = -\frac{1}{x - 2} \); \([-3, -2]\)
kuta software - infinite calculus
name_____________________________
average rates of change
date____________ period__
for each problem, find the average rate of change of the function over the given interval.
kuta software - infinite calculus
name_____________________________
average rates of change
date____________ period__
for each problem, find the average rate of change of the function over the given interval.
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 \( f(x) = x^2 - x + 1 \).
Substitute \( x = 0 \) into \( f(x) \):
\( f(0) = (0)^2 - 0 + 1 = 1 \).
Substitute \( x = 3 \) into \( f(x) \):
\( f(3) = (3)^2 - 3 + 1 = 9 - 3 + 1 = 7 \).
Using \(\frac{f(3) - f(0)}{3 - 0}\), substitute the values:
\(\frac{7 - 1}{3 - 0} = \frac{6}{3} = 2\).
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 \( f(x) = -\frac{1}{x - 2} \).
Substitute \( x = -3 \) into \( f(x) \):
\( f(-3) = -\frac{1}{-3 - 2} = -\frac{1}{-5} = \frac{1}{5} \).
Substitute \( x = -2 \) into \( f(x) \):
\( f(-2) = -\frac{1}{-2 - 2} = -\frac{1}{-4} = \frac{1}{4} \).
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, \(\frac{\frac{1}{20}}{1} = \frac{1}{20}\).
2
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 \( f(x) = x^2 - x + 1 \).
Substitute \( x = 0 \) into \( f(x) \):
\( f(0) = (0)^2 - 0 + 1 = 1 \).
Substitute \( x = 3 \) into \( f(x) \):
\( f(3) = (3)^2 - 3 + 1 = 9 - 3 + 1 = 7 \).
Using \(\frac{f(3) - f(0)}{3 - 0}\), substitute the values:
\(\frac{7 - 1}{3 - 0} = \frac{6}{3} = 2\).
2
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 \( f(x) = -\frac{1}{x - 2} \).
Substitute \( x = -3 \) into \( f(x) \):
\( f(-3) = -\frac{1}{-3 - 2} = -\frac{1}{-5} = \frac{1}{5} \).
Substitute \( x = -2 \) into \( f(x) \):
\( f(-2) = -\frac{1}{-2 - 2} = -\frac{1}{-4} = \frac{1}{4} \).
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, \(\frac{\frac{1}{20}}{1} = \frac{1}{20}\).
\(\frac{1}{20}\)
kuta software - infinite calculus
name_____________________________
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 )
Top-left cell: 180 Top-right cell: 6 Bottom-left cell: 600 Bottom-right cell: 20 Final product: 806
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