Question

Consider the function defined by f(x) lnx/x^4 for x > 0 and its graph y = f(x).

The graph of f has a horizontal tangent at point P. Find the coordinates of P.

Answer 1

To find the coordinates of point P where the graph of the function f(x) = ln(x)/x^4 has a horizontal tangent, we need to find the derivative of the function and set it equal to zero. The coordinates of point P are (1, f(1)).

Step-by-step explanation:

To find the coordinates of point P where the graph of the function f(x) = ln(x)/x^4 has a horizontal tangent, we need to find the derivative of the function and set it equal to zero. Let's start by finding the derivative.

First, apply the quotient rule to find the derivative of f(x):

f'(x) = [(x^4 * (1/x) - ln(x) * (4x^3))]/(x^4)^2

Simplify the expression:

f'(x) = [(x^3 - 4x^3 * ln(x))/x^4]/(x^8)

Now, set f'(x) equal to zero and solve for x:

[(x^3 - 4x^3 * ln(x))/x^4]/(x^8) = 0

x^3 - 4x^3 * ln(x) = 0

x^3(1 - 4ln(x)) = 0

Since x > 0, the solution to this equation is x = 1.

Therefore, the coordinates of point P are (1, f(1)).

Answer 2

`\sqrt[4]{e};\frac1{4e}`

Step-by-step explanation:

The first derivative of
`f(x)` will give, for each x, the slope of the tangent at that specific x. Let's calculate the derivative, applying the quotient rule.

`D(\frac AB) = (A'B-AB')/(B^2)\\f'(x)= \frac1{x^8}[(\frac1x)x^4-(lnx)(4x^3)]=\frac1{x^8}[x^3(1-4lnx)]=(1-4lnx)/(x^5)`

Now, to find the point with an horizontal tangent (called "stationary points"), we set the first derivative equal to 0. Considering that we're working with
`x > 0` we deal only with the numerator.

`1-4lnx = 0 \rightarrow lnx= \frac14\\x=e^\frac14 =\sqrt[4]{e}`

At this point we Replace the value we found in the equation to find it's y coordinate

`f(\sqrt[4]e) = (ln\sqrt[4]e)/(\sqrt[4]e^4)= (\frac14)/(e) = \frac1{4e}`