Solution by Steps
step 1
To find the average rate of change of f′ over the interval 3≤x≤12, we use the values of f′(x) at x=3 and x=12 step 2
The average rate of change is given by the formula x2−x1f′(x2)−f′(x1) step 3
Plugging in the values from the table, we get 12−3f′(12)−f′(3)=12−39−5 step 4
Simplifying, we find the average rate of change to be 94 [a] Answer
The average rate of change of f′ over the interval 3≤x≤12 is 94. Key Concept
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)) is incorrect as per the context given. The correct function is 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) and f′(x) are continuous and differentiable, m(x) is also continuous step 4
We need to check the signs of m(x) at the endpoints of the interval [3,12] step 5
Using the table, m(3)=f(3)−f′(3)=−2−5=−7 and m(12)=f(12)−f′(12)=15−9=6 step 6
Since m(3) is negative and m(12) is positive, there must be at least one value of x in [3,12] where m(x)=0 [b] Answer
There must be at least one value of x in [3,12] such that m(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], [5,12], and [12,15]. We use the values of f′′(x) at the right endpoints for the heights step 3
The approximation is ∑(f′′(xi)⋅Δxi) where xi are the right endpoints and Δxi are the widths of the subintervals step 4
Calculating, we get (−1⋅2)+(2⋅7)+(−6⋅3) step 5
Simplifying, the approximation is −2+14−18=−6 [c] Answer
The approximate value of the definite integral from 3 to 15 of f′′(x)dx using a right Riemann sum is −6. Key Concept
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), we need to evaluate the integral from −2 to 3⋅5=15 of f′(t)dt step 2
We can use the Fundamental Theorem of Calculus and the values of f(x) from the table since f′(t)dt is the derivative of f(t) step 3
The integral from −2 to 15 of f′(t)dt is f(15)−f(−2) step 4
Using the table, we get f(15)−f(−2)=36−(−10)=46 step 5
Therefore, g(5)=7+46=53 step 6
To find g′(5), we differentiate g(x) with respect to x step 7
By the Chain Rule, g′(x)=dxd(7+∫−23xf′(t)dt)=3f′(3x) step 8
Plugging in x=5, we get g′(5)=3f′(15) step 9
Using the table, f′(15)=2, so g′(5)=3⋅2=6 [d] Answer
g(5)=53 and g′(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.