The function ff is defined by f, of, x, equals, x, squared, minus, 3, x, minus, 2, cosine, left bracket, 2, x, squared, minus, 5, x, right bracket, .f(x)=x 2 −3x−2cos(2x 2 −5x). What is the average rate of change of f, primef ′ on the interval open square bracket, minus, 3, comma, 3, close square bracket, question mark[−3,3]? You may use a calculator and round your answer to the nearest thousandth
Question
The function ff is defined by f, of, x, equals, x, squared, minus, 3, x, minus, 2, cosine, left bracket, 2, x, squared, minus, 5, x, right bracket, .f(x)=x 2 −3x−2cos(2x 2 −5x). What is the average rate of change of f, primef ′ on the interval open square bracket, minus, 3, comma, 3, close square bracket, question mark[−3,3]? You may use a calculator and round your answer to the nearest thousandth
Solution
To find the average rate of change of the derivative of a function over an interval, we first need to find the derivative of the function.
The function is f(x) = x^2 - 3x - 2cos(2x^2 - 5x).
The derivative of this function, f'(x), can be found using the power rule, the constant rule, and the chain rule for derivatives.
The derivative of x^2 is 2x, the derivative of -3x is -3, and the derivative of -2cos(2x^2 - 5x) is 2sin(2x^2 - 5x) * (4x - 5) by the chain rule.
So, f'(x) = 2x - 3 + 2sin(2x^2 - 5x) * (4x - 5).
Next, we need to evaluate this derivative at the endpoints of the interval, which are -3 and 3.
f'(-3) and f'(3) can be found by plugging -3 and 3 into the derivative and simplifying.
Finally, the average rate of change of f' on the interval [-3,3] is given by the formula:
(f'(3) - f'(-3)) / (3 - (-3)).
This calculation can be done using a calculator, and the result should be rounded to the nearest thousandth.
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