Find the critical numbers of the function.
f(x) = x^-2ln(x)
A critical point is an interior point on the domain of a function f(x) where f'(x)=0 or f'(x) is undefined.
The domain of f(x)=x^(-2*ln(x)) is (0,∞) because ln(x) cannot take non-positive arguments.
Now, we need to find f'(x) and where f'(x)=0 or undefined.
If we take the logarithm on both sides and use implicit differentiation, we can find f'(x).
Let y=x^(-2ln(x))
ln(y)=-2ln(x)*ln(x)
ln(y)=-2(ln(x))²
differentiate with respect to x
(1/y)*(dy/dx)=-2(2ln(x))(1/x)
(dy/dx)=-4y ln(x)/x
f'(x)=-4x^(-2ln(x)) ln(x)/x
We see immediately that f'(x)=0 when x=1 (i.e. ln(x)=0).
Check to see if f'(x) is undefined at any point in the domain of f(x), and any other points where f'(x) = 0.
Check my work.
To find the critical numbers of a function, we need to identify the values of x where the derivative of the function is either zero or undefined.
First, let's find the derivative of the given function f(x) = x^(-2ln(x)) using the chain rule.
f'(x) = (-2ln(x)) * x^(-2ln(x)-1) * (1/x)
We can simplify this further by applying logarithmic and exponentiation properties:
f'(x) = (-2ln(x)) * (1/x) * e^(-2ln(x) ln(e))
Now, let's find where the derivative is either zero or undefined.
1. Undefined: The derivative is undefined when the denominator, x, equals zero. However, in this case, the function f(x) = x^(-2ln(x)) is not defined for x = 0. Therefore, we exclude x = 0 from the domain of the function.
2. Zero: To find where the derivative is equal to zero, we need to solve the equation f'(x) = 0. Since we have a product of three terms, we need to set each term equal to zero separately.
-2ln(x) = 0 --> ln(x) = 0 --> x = 1
1/x = 0 --> x = 0 (but we already know this is not defined)
e^(-2ln(x) ln(e)) = 0
The last term, e^(-2ln(x) ln(e)), cannot equal zero since the base of the exponential function (e) is a positive number and the exponent is negative. Therefore, there are no additional solutions for x.
In summary, the critical number of the function f(x) = x^(-2ln(x)) is x = 1.
To find the critical numbers of a function, we need to find the values of x where the derivative of the function is equal to zero or undefined.
Let's start by finding the derivative of f(x).
f(x) = x^(-2ln(x))
We can rewrite this as:
f(x) = e^(-2ln(x) * ln(x)^(-1))
Using the chain rule, the derivative is:
f'(x) = e^(-2ln(x) * ln(x)^(-1)) * ( -2 * ln(x)' * ln(x)^(-1) + 2 * ln(x) * ln(x)' * ln(x)^(-2) )
To simplify this, let's find the derivatives of ln(x) and ln(x)^(-1):
ln(x)' = 1/x
ln(x)^(-1)' = -1/(x * ln(x)^2)
Now let's substitute these derivatives back into the derivative of f(x):
f'(x) = e^(-2ln(x) * ln(x)^(-1)) * ( -2 * 1/x * ln(x)^(-1) + 2 * ln(x) * (-1/(x * ln(x)^2)) * ln(x)^(-2) )
Simplifying further:
f'(x) = e^(-2ln(x) * ln(x)^(-1)) * ( -2/x * ln(x)^(-1) - 2/(x^2 * ln(x)) * ln(x)^(-2) )
To find the critical numbers, we need to find where f'(x) = 0 or is undefined.
Setting f'(x) equal to zero:
-2/x * ln(x)^(-1) - 2/(x^2 * ln(x)) * ln(x)^(-2) = 0
Multiplying through by x^2 * ln(x),
-2 * ln(x)^(-1) - 2 * ln(x)^(-2) = 0
Simplifying,
-2 * ln(x)^(-1) - 2/ln(x) = 0
Multiplying through by ln(x),
-2 - 2/ln(x)^2 = 0
Simplifying further,
-2 * ln(x)^2 - 2 = 0
Dividing through by -2,
ln(x)^2 + 1 = 0
Since the natural logarithm of a number can never be negative, there are no solutions to this equation. Therefore, there are no critical numbers for the function f(x) = x^(-2ln(x)).