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- Find the average rates of change of f(x)=x2+2x (a) from x1=3 to x2=2 and (b) from x1=2 to x2=0.find the point(s) that are guaranteed to exist by the Mean Value Theorem. In his problem the Mean Value Theorem may not apply. If so justify why. F(x) = 7 – x2 ; [-1,2]If f(x) = |x|, then f(−1) = 1 and f(3) = 3 but f'(x) is never equal to f(3) − f(−1)/3 − (−1)=1/2. Why doesn’t this violate the Mean Value Theorem?
- Consider the function f(x)=1/x on the interval [3,12]. Find the average or mean slope of the function on this interval. =? By the Mean Value Theorem, we know there exists a c in the open interval (3,12) such that f′(c) is equal to this mean slope. For this problem, there is only one c that works. Find it. =?Find the mean value of the function f(x)=|8−x| on the closed interval [7,9].(TRUE / FALSE) If θ^MM is the MM estimator of θ, f(θ^MM ) is the MM estimator of f(θ) where f is a continuous function.
- If f(x) = |x|, then f(−1) = 1 and f(3) = 3 butf(x) is never equal to f(3) − f(−1)/3 − (−1)=1/2. Whydoesn’t this violate the Mean Value Theorem?Consider the function f(x)=7-5x2 on the interval [-2,5]. Find the average or mean slope of the function on this interval, i.e.f(5)−f(−2)/5−(−2)= ? By the Mean Value Theorem, we know there exists a c in the open interval (-2,5) such that f′(c) is equal to this mean slope. For this problem, there is only one cc that works. Find it. = ?Assume that the entire sample has 2 positive observations and 6 negatives observations. Variable X1: at the left branch has 2 positive observations and 4 negatives observations;at the right branch has 6 positive observations and 6 negatives observations. What is the information gain or reduction in uncertainty of X1 using the Gini index? (round to two decimal spaces) I am getting negative value -1.37 for this question.But information gain should be positive.
- Consider the function f(x)=3x^3−2x on the interval [-2,2]. Find the average or mean slope of the function on this interval? =? By the Mean Value Theorem, we know there exists at least one c in the open interval (-2,2) such that f'(c) is equal to this mean slope. For this problem, there are two values of c that work. The smaller one is =? and the larger one is =?Q1)The function f(x)=x(x-6)^2satisfies both hypotheses of the Mean Value Theorem on the interval [0,8],Determine the slope of the secant line connecting these two points and find all values of c on that interval guaranteed by the conclusion of the MVT.The Mean Value Theorem for problem 3 says that between x=-2 and x=2, there exists a value x=c that f'(c) = f(2) -f(-2)/2-(-2). Find the actual value for c which makes the MVT true in this case. Function for problem 3: y=27-x2