Find constants a and b in the function f(x)=(ax^b)ln(x) such that f(1/5)=1 and the function has a local minimum at x=1/5.
f = ax^b lnx
f' = ax^b/x + ax^b lnx b/x
= ax^blnx (1/xlnx + b/x)
= f(x)*(1/xlnx + b/x)
f'(1/5) = 1/(1/5 ln(1/5)) + 5b
= -5/ln5 + 5b
so, b = 1/ln5
f(x) = ax^(1/ln5) lnx
a(1/5)^(1/ln5)(-ln5) = 1
a = -5^(1/ln5) / ln5 = -e/ln5
f(x) = -e/ln5 x^(1/ln5) lnx
This has a minimum at x = 1/5, but f(1/5) = -1.
On the other hand, with
f(x) = e/ln5 x^(1/ln5) lnx
f(1/5) = 1, but it has a local max there.
Better check for typos, and double-check my math.
See
http://www.wolframalpha.com/input/?i=plot+y%3De%2Fln5+x%5E%281%2Fln5%29+lnx+for+x%3D0..1
To find the constants a and b in the function f(x) = (ax^b)ln(x), given that f(1/5) = 1 and the function has a local minimum at x = 1/5, we can set up the following equations:
Equation 1: f(1/5) = 1
Equation 2: f'(1/5) = 0 (to find the local minimum)
Let's start with Equation 1:
Substituting x = 1/5 into the function f(x), we have:
f(1/5) = (a(1/5)^b)ln(1/5) = 1
Now let's work on Equation 2 to find the derivative of f(x):
Taking the derivative, f'(x), of f(x) = (ax^b)ln(x) using the product rule, we get:
f'(x) = (a(ln(x) + 1))x^b + (ax^b)(1/x)
Setting x = 1/5:
f'(1/5) = (a(ln(1/5) + 1))(1/5)^b + (a(1/5)^b)(1/(1/5))
= (a(ln(1/5) + 1))(1/5)^b + 5a(1/5)^b
Now, let's solve the two equations simultaneously:
From Equation 1, we have:
(a(1/5)^b)ln(1/5) = 1
Since ln(1/5) = -ln(5), we can rewrite the equation as:
(a(1/5)^b)(-ln(5)) = 1
Dividing both sides by (-ln(5)), we get:
a(1/5)^b = -1 / ln(5)
Now, let's substitute this value into Equation 2:
(a(ln(1/5) + 1))(1/5)^b + 5a(1/5)^b = 0
Substituting -1 / ln(5) for a(1/5)^b, we have:
(-1 / ln(5))(ln(1/5) + 1) + 5(-1 / ln(5)) = 0
Simplifying the equation:
(-ln(1/5) - ln(5) + ln(5)) / ln(5) - 5 / ln(5) = 0
(-ln(1/5)) / ln(5) - 5 / ln(5) = 0
-ln(1/5) - 5 = 0
-ln(1/5) = 5
ln(1/5) = -5
Therefore, there is no solution for constants a and b such that f(1/5) = 1 and the function has a local minimum at x = 1/5.
To find the constants a and b in the function f(x) = (ax^b)ln(x) such that f(1/5) = 1 and the function has a local minimum at x = 1/5, we will use the given information and the properties of local minimum.
First, let's substitute the values into the function:
f(1/5) = (a(1/5)^b)ln(1/5) = 1
This equation gives us the first condition that we need to solve. Now, we'll use the fact that the function has a local minimum at x = 1/5. For any value of x, the function has a local minimum when its derivative is zero and the second derivative is positive.
Let's find the derivative of f(x):
f'(x) = d/dx[(ax^b)ln(x)]
= (ax^b)(d/dx[ln(x)]) + ln(x)(d/dx[ax^b])
= (ax^b)(1/x) + ln(x)(abx^(b-1))
Simplifying further:
f'(x) = a(x^(b-1)) + ab(x^b)ln(x)
Now, let's find the second derivative:
f''(x) = d/dx[f'(x)]
= d/dx[a(x^(b-1)) + ab(x^b)ln(x)]
= ab(x^(b-1))(d/dx[ln(x)]) + abln(x)(d/dx[x^b])
= ab(x^(b-1))(1/x) + abln(x)(bx^(b-1))
Simplifying further:
f''(x) = ab(x^(b-2)) + ab^2x^(b-1)ln(x)
Since we know that x = 1/5 is a local minimum, we can set f'(1/5) = 0 and f''(1/5) > 0.
Setting x = 1/5 in f'(x) and f''(x), we get:
f'(1/5) = a((1/5)^(b-1)) + ab((1/5)^b)ln(1/5) = 0
f''(1/5) = ab((1/5)^(b-2)) + ab^2(1/5)^(b-1)ln(1/5)
Now we have two equations:
1. a((1/5)^(b-1)) + ab((1/5)^b)ln(1/5) = 0
2. ab((1/5)^(b-2)) + ab^2(1/5)^(b-1)ln(1/5) > 0
To solve these equations, we need to make some assumptions about the values of a and b. Let's assume a ≠ 0 and b ≠ 0.
First, let's solve the equation f(1/5) = 1 to find the value of a.
(a(1/5)^b)ln(1/5) = 1
a(1/5)^b = 1/ln(1/5)
a = (1/ln(1/5)) * (5^b)
Now, we substitute this value of a into the equations:
((1/ln(1/5)) * (5^b) * (1/5)^(b-1)) + (((1/ln(1/5)) * (5^b)) * (1/5)^b)ln(1/5) = 0
((1/ln(1/5)) * (5^b) * (1/5)^(b-2)) + (((1/ln(1/5)) * (5^b)) * (1/5)^(b-1)) * b * ln(1/5) > 0
At this point, the equations may become more complicated to solve explicitly due to the appearance of the logarithm and fractional exponents. A numerical approach or approximation technique may be required to find suitable values for a and b that satisfy the equations and the given conditions.
Note: The natural logarithm (ln) function is typically used in these calculations, but other logarithmic bases can be used as well.