find the critical no's/ points in
(X^-2)(lnx)
let y = (x^-2)lnx
dy/dx = lnx(-2x^-3) + x^-2(1/x)
= -2lnx/x^3 + 1/x^3 = 0 for max/min of y
multiply by x^3
2lnx = 1
lnx = 1/2
x = e^(1/2) or √e
then y = ln(√e)/e = 1/(2e)
critical point is (√e, 1/(2e))
hey thnxx but web assign is not accepting this ans, i dunno why it says that its wrong.
try e^(1/2) instead of √e or find the decimals
the above point in rounded decimals is
(1.6487, .1839)
its wrong again :(
To find the critical points or critical numbers of the function f(x) = x^(-2) * ln(x), we need to determine where the derivative of the function is equal to zero or undefined.
Let's start by finding the derivative of the function f(x):
f'(x) = d/dx [x^(-2) * ln(x)]
To simplify this expression, we can use the product rule:
f'(x) = (d/dx [x^(-2)]) * ln(x) + x^(-2) * (d/dx [ln(x)])
To find the derivative of x^(-2), we can use the power rule:
d/dx [x^(-2)] = -2x^(-3)
To find the derivative of ln(x), we can use the chain rule:
d/dx [ln(x)] = 1/x
Substituting these derivatives back into the expression for f'(x):
f'(x) = (-2x^(-3)) * ln(x) + x^(-2) * (1/x)
Simplifying this expression:
f'(x) = -2/x^3 * ln(x) + 1/x^3
To find the critical values, we need to solve the equation f'(x) = 0:
-2/x^3 * ln(x) + 1/x^3 = 0
Multiplying both sides by x^3:
-2 * ln(x) + 1 = 0
Now, we can isolate the term ln(x):
-2 * ln(x) = -1
Dividing both sides by -2:
ln(x) = 1/2
Exponentiating both sides with base e:
e^(ln(x)) = e^(1/2)
x = e^(1/2)
Therefore, x = e^(1/2) is a critical value of the function.
However, we also need to check if there are any values of x that make the derivative undefined. Since our derivative involves ln(x), x must be greater than 0. Therefore, x = 0 is not a valid point to consider.
In conclusion, the critical value or critical point of the function f(x) = x^(-2) * ln(x) is x = e^(1/2), given that x > 0.