Find constants a and b in the equation f(x)= ax^b/(ln(x)) such that f(1/3)= 1 and the function has a local minimum at x= 1/3
To find constants a and b in the equation f(x) = ax^b / (ln(x)) such that f(1/3) = 1 and the function has a local minimum at x = 1/3, we need to use two conditions: f(1/3) = 1 and the derivative of f(x) at x = 1/3 is equal to 0.
First, let's find f(1/3) = 1 by plugging in x = 1/3 in the equation:
f(1/3) = a * (1/3)^b / ln(1/3) = 1
Now, let's find the derivative of f(x) with respect to x:
f'(x) = d/dx [ ax^b / ln(x) ]
To do this, we will use the quotient rule.
Using the quotient rule, the derivative of f(x) is:
f'(x) = [b * ax^(b-1) * ln(x) - ax^b * (1/x)] / (ln(x))^2
Now, let's calculate f'(1/3) and set it equal to 0 since we want a local minimum at x = 1/3:
f'(1/3) = [b * a * (1/3)^(b-1) * ln(1/3) - a * (1/3)^b * (1/(1/3))] / (ln(1/3))^2
Setting f'(1/3) = 0, we have:
[b * a * (1/3)^(b-1) * ln(1/3) - 3a * (1/3)^b] / (ln(1/3))^2 = 0
Now, we have two equations:
1) a * (1/3)^b / ln(1/3) = 1
2) [b * a * (1/3)^(b-1) * ln(1/3) - 3a * (1/3)^b] / (ln(1/3))^2 = 0
From equation 1, we can simplify it to:
a / ln(1/3) = 3^b
Substituting this into equation 2, we have:
[b * 3^b - 3a] / (ln(1/3))^2 = 0
At this point, we have a system of equations:
a / ln(1/3) = 3^b
[b * 3^b - 3a] / (ln(1/3))^2 = 0
By solving this system of equations, we can find the values of a and b that satisfy the given conditions. However, finding the exact values of a and b requires further calculations and might involve numerical methods like Newton's method or using a computer algebra system.