Find constants a and b in the equation f(x)= ax^b/(ln(x)) such that f(1/8)= 1 and the function has a local minimum at x= 1/8
a=-0.7650
b=-0.4809
I concur
To find the constants a and b in the equation f(x) = ax^b/(ln(x)), such that f(1/8) = 1 and the function has a local minimum at x = 1/8, we can use the given information to set up a system of equations.
We know that f(1/8) = 1, so let's substitute the values into the equation:
1 = a(1/8)^b / ln(1/8)
We also know that the function has a local minimum at x = 1/8, which means its derivative is zero at that point. The derivative of the given function is:
f'(x) = (a*b*x^(b-1)*ln(x) - a*x^b) / (x*ln^2(x))
Since the function has a local minimum at x = 1/8, we can set the derivative equal to zero and solve for a and b:
0 = (a*b*(1/8)^(b-1)*ln(1/8) - a*(1/8)^b) / ((1/8)*ln^2(1/8))
Let's simplify this equation:
0 = a*b*(1/8)^(b-1)*ln(1/8) - a*(1/8)^b
Now, we have a system of two equations:
1 = a(1/8)^b / ln(1/8)
0 = a*b*(1/8)^(b-1)*ln(1/8) - a*(1/8)^b
To solve this system, we can use algebraic methods or numerical methods such as Newton's method or the bisection method to find the values of a and b that satisfy both equations.