Consider the following function.

Question

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.)

Answer 1

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.