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\begin{tabular}{|c|c|c|c|c|c|} \hlinexx & -2 & 3 & 5 & 12 & 15 \\ \hlinef(x)f(x)...
May 10, 2024
Solution by Steps
step 1
To find the average rate of change of ff' over the interval 3x123 \leq x \leq 12, we use the values of f(x)f'(x) at x=3x=3 and x=12x=12
step 2
The average rate of change is given by the formula f(x2)f(x1)x2x1\frac{f'(x_2) - f'(x_1)}{x_2 - x_1}
step 3
Plugging in the values from the table, we get f(12)f(3)123=95123\frac{f'(12) - f'(3)}{12 - 3} = \frac{9 - 5}{12 - 3}
step 4
Simplifying, we find the average rate of change to be 49\frac{4}{9}
[a] Answer
The average rate of change of ff' over the interval 3x123 \leq x \leq 12 is 49\frac{4}{9}.
Key Concept
Average Rate of Change
Explanation
The average rate of change of a function over an interval is the change in the function values divided by the change in the input values over that interval.
step 1
The function m(x)=f(f(x))m(x) = f(f(x)) is incorrect as per the context given. The correct function is m(x)=f(x)f(x)m(x) = f(x) - f'(x)
step 2
By the Intermediate Value Theorem, if a continuous function changes signs over an interval, it must cross zero at least once within that interval
step 3
Since f(x)f(x) and f(x)f'(x) are continuous and differentiable, m(x)m(x) is also continuous
step 4
We need to check the signs of m(x)m(x) at the endpoints of the interval [3,12][3, 12]
step 5
Using the table, m(3)=f(3)f(3)=25=7m(3) = f(3) - f'(3) = -2 - 5 = -7 and m(12)=f(12)f(12)=159=6m(12) = f(12) - f'(12) = 15 - 9 = 6
step 6
Since m(3)m(3) is negative and m(12)m(12) is positive, there must be at least one value of xx in [3,12][3, 12] where m(x)=0m(x) = 0
[b] Answer
There must be at least one value of xx in [3,12][3, 12] such that m(x)=0m(x) = 0.
Key Concept
Intermediate Value Theorem
Explanation
The Intermediate Value Theorem states that if a function is continuous on a closed interval and takes on different signs at the endpoints, then it must have a zero within that interval.
step 1
To approximate the integral using a right Riemann sum, we multiply the height of each subinterval by the width of that subinterval
step 2
The subintervals are [3,5][3,5], [5,12][5,12], and [12,15][12,15]. We use the values of f(x)f''(x) at the right endpoints for the heights
step 3
The approximation is (f(xi)Δxi)\sum (f''(x_i) \cdot \Delta x_i) where xix_i are the right endpoints and Δxi\Delta x_i are the widths of the subintervals
step 4
Calculating, we get (12)+(27)+(63)(-1 \cdot 2) + (2 \cdot 7) + (-6 \cdot 3)
step 5
Simplifying, the approximation is 2+1418=6-2 + 14 - 18 = -6
[c] Answer
The approximate value of the definite integral from 3 to 15 of f(x)dxf''(x) dx using a right Riemann sum is 6-6.
Key Concept
Right Riemann Sum
Explanation
A right Riemann sum approximates the value of a definite integral by summing the products of the function values at the right endpoints of subintervals and the widths of those subintervals.
step 1
To find g(5)g(5), we need to evaluate the integral from 2-2 to 35=153 \cdot 5 = 15 of f(t)dtf'(t) dt
step 2
We can use the Fundamental Theorem of Calculus and the values of f(x)f(x) from the table since f(t)dtf'(t) dt is the derivative of f(t)f(t)
step 3
The integral from 2-2 to 1515 of f(t)dtf'(t) dt is f(15)f(2)f(15) - f(-2)
step 4
Using the table, we get f(15)f(2)=36(10)=46f(15) - f(-2) = 36 - (-10) = 46
step 5
Therefore, g(5)=7+46=53g(5) = 7 + 46 = 53
step 6
To find g(5)g'(5), we differentiate g(x)g(x) with respect to xx
step 7
By the Chain Rule, g(x)=ddx(7+23xf(t)dt)=3f(3x)g'(x) = \frac{d}{dx} \left(7 + \int_{-2}^{3x} f'(t) dt\right) = 3f'(3x)
step 8
Plugging in x=5x=5, we get g(5)=3f(15)g'(5) = 3f'(15)
step 9
Using the table, f(15)=2f'(15) = 2, so g(5)=32=6g'(5) = 3 \cdot 2 = 6
[d] Answer
g(5)=53g(5) = 53 and g(5)=6g'(5) = 6.
Key Concept
Fundamental Theorem of Calculus and Chain Rule
Explanation
The Fundamental Theorem of Calculus relates the integral of a function to its antiderivative, and the Chain Rule is used to differentiate composite functions.
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