Let (2,3) be the largest open interval in which the function f(x)=2loge(x−2)−x2+ax+1 is strictly increasing and (b, c) be the largest open interval in which the function g(x)=(x−1)3(x+2−a)2 is strictly decreasing. Then 100(a + b - c) is equal to:
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Step-by-Step Solution
Step 1: Find the derivative of f(x) and determine 'a'
To find where the function f(x) is strictly increasing, we need to find its first derivative, f'(x), and set it greater than zero. The derivative of loge(x−2) is x−21, and the derivative of −x2+ax+1 is -2x + a. Since f(x) is strictly increasing in (2,3), f'(x) > 0 for x∈(2,3). The critical points occur when f′(x)=0. Given that (2,3) is the largest interval where f(x) is strictly increasing, x=3 must be a critical point, meaning f′(3)=0.
Step 2: Solve for 'a'
Substitute x=3 into the derivative f'(x) and set it equal to zero to solve for the constant a. This gives us the value of a that makes x=3 a critical point.
Step 3: Find the derivative of g(x)
Next, we need to find the derivative of g(x) to determine where it is strictly decreasing. We use the product rule for differentiation. Substitute a=4 into the expression for g(x) before or after differentiation.
Step 4: Simplify g'(x) and find critical points
Factor out common terms from g'(x) to simplify it. Then substitute a=4 into the simplified expression. The critical points are where g′(x)=0.
Step 5: Determine the intervals for g(x) decreasing
Substitute a=4 into the factored form of g'(x) and simplify the expression. The critical points are x=1, x=2, and x=58=1.6. We need to analyze the sign of g'(x) in the intervals defined by these critical points. Since (x−1)2≥0, the sign of g'(x) depends on (x-2)(5x-8). g(x) is strictly decreasing when g'(x) < 0. This occurs when (x−2) and (5x−8) have opposite signs, which is for x∈(58,2). Thus, b=58 and c=2.
Step 6: Calculate the final value
Finally, substitute the values of a, b, and c into the expression 100(a + b - c) and calculate the result.