Consider the following function.

f(x) = x^6 ln x

(a) Use l'Hospital's Rule to determine the limit as x → 0+.

(b) Use calculus to find the minimum value.

Find the interval where the function is concave up. (Enter your answer in interval notation.)

Find the interval where the function is concave down. (Enter your answer in interval notation.)

basic stuff.

f(x) = lnx/x^-6
limit is same as (1/x)/-6x^-5 = -x^4/6 = 0

f'(x) = x^6 * 1/x + 6x^5 lnx = x^5 (1+6lnx)
so, f'=0 when x = e^(-1/6)

f" = x^4 (11+30lnx)
x^4 is always positive, so, f" > 0 when lnx > -11/30, or x > e^(-11/30)

I expect you can rephrase that as an interval.

(a) To find the limit as x → 0+ using l'Hospital's Rule, we need to find the limit of the derivative of the function as x approaches 0+.

f'(x) = 6x^5 ln x + x^5 / x = 6x^5 ln x + x^4

Now find the limit of f'(x) as x approaches 0+:

lim(x→0+) f'(x) = lim(x→0+) [6x^5 ln x + x^4]

Using l'Hospital's Rule again, let's find the limit of the derivative of f'(x):

f''(x) = 30x^4 ln x + 10x^3 / x = 30x^4 ln x + 10x^2

lim(x→0+) f''(x) = lim(x→0+) [30x^4 ln x + 10x^2]

Using l'Hospital's Rule for the third time, we find:

f'''(x) = 120x^3 ln x + 20x / x = 120x^3 ln x + 20

lim(x→0+) f'''(x) = lim(x→0+) [120x^3 ln x + 20]

Using l'Hospital's Rule once again:

f''''(x) = 360x^2 ln x + 20 / x = 360x^2 ln x + 20

lim(x→0+) f''''(x) = lim(x→0+) [360x^2 ln x + 20]

Continuing this process, we eventually get down to:

f''''''(x) = 720x ln x / x = 720x ln x

lim(x→0+) f''''''(x) = lim(x→0+) [720x ln x]

Now we can evaluate the limit as x approaches 0+:

lim(x→0+) f''''''(x) = lim(x→0+) [720x ln x] = 0

Therefore, the limit as x → 0+ of the function f(x) = x^6 ln x is 0.

(b) To find the minimum value of the function, we take the derivative of f(x) and set it equal to 0:

f'(x) = 6x^5 ln x + x^4 = 0

We need to solve this equation to find the critical points. Since ln x cannot be zero, we can ignore the first term and solve for x:

x^4 = 0

The only solution to this equation is x = 0. Therefore, x = 0 is a critical point.

Next, we check the behavior of f(x) to determine if it's a minimum or maximum. We can do this using the second derivative test:

f''(x) = 30x^4 ln x + 10x^2

Now substitute x = 0 into the second derivative:

f''(0) = 0^4 ln 0 + 10(0)^2 = 0

Since the second derivative is equal to 0 at x = 0, the second derivative test is inconclusive. In this case, we need to consider other methods to determine if the critical point is a minimum or maximum.

To do this, we can use the first derivative test or check the behavior of the function for values close to x = 0. From the first derivative equation, we can see that as x approaches 0+, the first term becomes negligible compared to the second term.

Therefore, f'(x) > 0 for x > 0 and f'(x) < 0 for x < 0.

This implies f(x) is increasing for x > 0 and decreasing for x < 0, meaning the critical point x = 0 is a minimum.

For the interval where the function is concave up, we need to find where the second derivative is positive:

f''(x) > 0

30x^4 ln x + 10x^2 > 0

Since the natural logarithm (ln x) is always positive for positive x, we only need to consider the second term:

10x^2 > 0

This inequality is true for all x ≠ 0, so the function is concave up for all x ≠ 0.

In interval notation, the interval where the function is concave up is (-∞, 0) U (0, +∞).

For the interval where the function is concave down, we need to find where the second derivative is negative:

f''(x) < 0

30x^4 ln x + 10x^2 < 0

Again, we consider only the second term:

10x^2 < 0

This inequality is true for x < 0, so the function is concave down for x < 0.

In interval notation, the interval where the function is concave down is (-∞, 0).

To solve these problems, we will go step by step.

(a) To determine the limit as x approaches 0+ using l'Hospital's Rule, follow these steps:

1. Calculate the derivative of the numerator and denominator of the function f(x) = x^6 ln x. The derivative of x^6 ln x is 6x^5 * ln x + x^6 * (1/x) = 6x^5 ln x + x^5.

2. Evaluate the limit of the derivative of the function as x approaches 0+. We have:

lim (x → 0+) [6x^5 ln x + x^5].

At this point, we can substitute x = 0 to check if the limit is indeterminate (0/0) or not. If it is, we can apply l'Hospital's Rule again. However, in this case, plugging in x = 0 gives us [0 * ln 0 + 0] = 0.

Therefore, the limit as x approaches 0+ is 0.

(b) To find the minimum value of the function f(x) = x^6 ln x using calculus, follow these steps:

1. Take the derivative of the function f(x) with respect to x:

f'(x) = d/dx (x^6 ln x) = 6x^5 ln x + x^6 * (1/x) = 6x^5 ln x + x^5.

2. Set the derivative equal to zero and solve for x to find critical points:

6x^5 ln x + x^5 = 0.

We can factor out x^5:

x^5(6 ln x + 1) = 0.

This equation is satisfied when either x^5 = 0 or 6 ln x + 1 = 0.

The first equation, x^5 = 0, has a single solution at x = 0.

For the second equation, 6 ln x + 1 = 0, we can solve for x:

6 ln x = -1,
ln x = -1/6,
x = e^(-1/6).

Therefore, the critical points are x = 0 and x = e^(-1/6).

3. Use the second derivative test to determine if the critical points are local minima or maxima.

To do this, we need to find the second derivative of the function:

f''(x) = d^2/dx^2 (x^6 ln x) = 30x^4 ln x + x^4.

Now substitute x = e^(-1/6) into f''(x):

f''(e^(-1/6)) = 30(e^(-1/6))^4 ln (e^(-1/6)) + (e^(-1/6))^4,

Simplifying gives:

f''(e^(-1/6)) = 30e^(-4/6) ln e^(-1/6) + e^(-4/6),

Since ln e^(-1/6) = -1/6, we have:

f''(e^(-1/6)) = 30e^(-4/6) (-1/6) + e^(-4/6),

Simplifying further:

f''(e^(-1/6)) = (-5/6)e^(-4/6) + e^(-4/6) = -1/6 e^(-4/6).

Since e^(-4/6) is positive, and the coefficient of e^(-4/6) is negative, f''(e^(-1/6)) is negative. Therefore, x = e^(-1/6) is a local maximum.

4. To determine if x = 0 is a local minimum or maximum, we can substitute values near x = 0 into the second derivative:

f''(0) = 0^4 ln 0 + 0^4 = 0.

Since the second derivative at x = 0 is 0, the second derivative test is inconclusive, and we need to analyze the function using other methods.

By observing the behavior of the function as x approaches 0 from both sides, we see that the function's values are negative for x > 0 and approach 0 as x approaches 0+.

Therefore, x = 0 corresponds to a local minimum value of 0.

The interval where the function is concave up can be found by examining the sign of the second derivative:

f''(x) = 30x^4 ln x + x^4.

For the function to be concave up, f''(x) needs to be positive.

We know that 0 ≤ x < e^(-1/6).

Substituting x = 1 into f''(x):

f''(1) = 30(1)^4 ln 1 + (1)^4 = 1,

Therefore, the function is concave up on the interval (0, e^(-1/6)).

The interval where the function is concave down can be found by examining the sign of the second derivative:

f''(x) = 30x^4 ln x + x^4.

For the function to be concave down, f''(x) needs to be negative.

We know that 0 ≤ x < e^(-1/6).

Substituting x = 2 into f''(x):

f''(2) = 30(2)^4 ln 2 + (2)^4 = 64 + 120 ln 2,

Since ln 2 is positive, f''(2) > 0.

Therefore, the function is concave down on the interval [e^(-1/6), ∞).