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To solve this problem, we must first differentiate:
Identify that we are able to use the product rule as our expression is of the form y = f(x)g(x) where f(x) = e^(2- x) and g(x) = ln(3x- 2).
Hence f'(x) = -e^(2- x) and g'(x) = 3/(3x- 2)
By the product rule, dy/dx = f(x)g'(x) + f'(x)g(x) = 3e^(2- x)/(3x- 2) - e^(2- x)ln(3x- 2).
When we substitute x = 2 into this equation, we get that dy/dx = 3/4 - ln(4), which is our final answer.see more
In order to deal with the modulus sign, we must take account of 2 possible cases:
Case 1: |2x + 1| = (2x +1). In this case we can solve algebraicly, preserving the inequality sign, to get that x < 4 -(2x + 1) = 3 - 2x. Then by adding 2x to each side and dividing both sides by 3 we get x < 1.
Case 2: |2x + 1| = -(2x +1). In this case we solve algebraicly again so that x < 4 + (2x +1) = 2x + 5. Hence by subtracting a 5 and an x from each side we get x > -5.
Finally we combine the results of each case, namely that x < 1 and x > -5 to get -5 < x < 1 as our final solution.see more