Find constants a and b in the function f(x)=axbln(x) such that f(15)=1 and the function has a local minimum at x=15.
To find constants a and b in the function f(x) = ax^b ln(x) such that f(15) = 1 and the function has a local minimum at x = 15, we need to use two conditions: f(15) = 1 and the derivative of f(x) at x = 15 is equal to zero.
Let's start by evaluating f(15):
f(15) = a * 15^b * ln(15)
We are given that f(15) = 1, so we can write:
1 = a * 15^b * ln(15) -- (Equation 1)
Next, we need to find the derivative of f(x) and set it equal to zero to find the value of b. Let's find the derivative of f(x):
f'(x) = D(a * x^b * ln(x))
Using the product rule and the chain rule of differentiation, we have:
f'(x) = a * b * x^(b-1) * ln(x) + a * x^b * 1/x
We want to find the value of b such that f'(15) = 0. Let's substitute x = 15 into f'(x):
f'(15) = a * b * 15^(b-1) * ln(15) + a * 15^b * 1/15
We can simplify this equation further:
0 = a * b * 15^(b-1) * ln(15) + a * 15^(b-1)
Dividing both sides of the equation by a * 15^(b-1), we get:
0 = b * ln(15) + 1
Subtracting 1 from both sides, we have:
-1 = b * ln(15)
Finally, to solve for b, divide both sides of the equation by ln(15):
b = -1 / ln(15)
Now that we have found the value of b, we can substitute it back into Equation 1 to solve for a. Recall Equation 1:
1 = a * 15^b * ln(15)
We substitute b = -1 / ln(15):
1 = a * 15^(-1 / ln(15)) * ln(15)
Now, solving for a:
a = 1 / (15^(-1 / ln(15)) * ln(15))
Using a calculator, we can evaluate a:
a ≈ 0.318376
So, the constants a and b that satisfy the given conditions are approximately:
a ≈ 0.318376
b ≈ -1 / ln(15)