Consider the function f(x)=ln(x)/x^6. For this function there are two important intervals: (A,B] and [B,∞) where A and B are critical numbers or numbers where the function is undefined.
Find A
Find B
For each of the following intervals, tell whether f(x) is increasing or decreasing .
(A,B]:
[B,∞):
I know how to take the derivative the other parts i don't know how to do...
A: Hmm, finding critical numbers can be a bit of a puzzle. Let me put on my thinking wig...oh wait, I don't have hair. Anyway, let's figure this out. To find A, we need to find where the function is undefined or where the denominator becomes zero. In this case, x^6 cannot be zero, so A is not defined. It's like trying to divide by zero, we'll just have to let it be a mystery.
B: Now for B, we need to find where the function is undefined or where the denominator becomes zero. In this case, we need to solve the equation x^6 = 0. But wait a second, x raised to any power will never be zero unless x itself is zero. So B = 0 because f(x) becomes undefined at x = 0.
For (A, B]: Since A is undefined, it's like finding the color of a unicorn's horn, so we don't have a valid interval here.
For [B, ∞): Now that we know B = 0, let's analyze this interval. Well, for positive values of x, the numerator ln(x) will always be positive. And since x^6 is also positive, the function f(x) will always be positive on this interval. So, f(x) is increasing since laughter always goes up!
Hope that clears things up!
To find the critical numbers or numbers where the function is undefined for the given function f(x) = ln(x)/x^6, we need to look for values of x where the function is either undefined or has zero derivative.
First, let's determine the values of x where the function is undefined. Since ln(x) is defined only for positive values of x, we need to find the values of x where x^6 is equal to 0. However, x^6 can never be equal to 0 since any number raised to the power of 6 will always be positive. Therefore, the function is undefined for any x ≤ 0.
Next, let's find the values of x where the derivative of the function is zero or does not exist. To do this, we need to find the critical points of the function by taking the derivative and solving for x.
The derivative of f(x) is given by:
f'(x) = (1/x) * (1/x^6) - ln(x) * 6/x^7
= 1/(x^7) - 6ln(x)/(x^7)
= (1 - 6ln(x)) / (x^7)
To find the critical points, we need to solve for x when the derivative is zero:
1 - 6ln(x) = 0
Rearranging the equation, we have:
6ln(x) = 1
ln(x) = 1/6
Taking the exponential of both sides, we get:
x = e^(1/6)
Therefore, the critical number A is x = e^(1/6), which is approximately 1.180.
Now, to find the second important interval [B, ∞), we need to find the largest possible value of x for the function. Since x approaches infinity, the interval [B, ∞) is defined where x > e^(1/6). Therefore, B = e^(1/6).
To determine whether f(x) is increasing or decreasing on the intervals (A, B] and [B, ∞), we can examine the sign of the derivative, f'(x).
On the interval (A, B], the derivative f'(x) = (1 - 6ln(x)) / (x^7) is always positive because x > e^(1/6). Therefore, f(x) is increasing on (A, B].
On the interval [B, ∞), the derivative f'(x) is always negative because x > e^(1/6). Therefore, f(x) is decreasing on [B, ∞).
In summary:
A = e^(1/6) (approximately 1.180)
B = e^(1/6) (approximately 1.180)
Interval (A,B]: f(x) is increasing.
Interval [B,∞): f(x) is decreasing.
I will give you the derivative, you should be able to take it from there
f ' (x) = (x^6(1/x) - lnx(6x^5) )/x^12
= (x5 - 6x^5(lnx))/x^12