Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 16

f(x) = (x^2)ln(x)
(a) Find the intervals on which f is increasing or decreasing. (b) Find the local maximum and minimum values of f. (c) Find the intervals of concavity and the inflection points
f(x)=(x^2)ln(x)
(a) Take the derivative of the function.
f'(x) =x^2 * 1/x + 2x * ln(x)
f'(x)= x +2xln(x)
Set the derivative equal to zero.
0=x+2xln(x)
Factor the right side.
0=x(1 + 2ln(x))
Then, set each factor to zero and isolate the x.
For the first factor:
x = 0
For the second factor:
1 + 2lnx = 0
2lnx=-1
lnx = -1/2
x =e^(-1/2) Hence, the critical numbers are x=0 and x=e^(-1/2) .
To get the intervals formed by these two critical numbers, we have to consider the domain of the function f(x). Take note that in logarithm, zero and negative numbers are not allowed. So the domain of f(x) is (0, oo) .
Considering the domain of the function, the intervals formed by the critical number are (0, e^(-1/2)) and (e^(-1/2),oo) .
Then, assign a test value for each interval. Plug-in them to the first derivative.
f'(x) = x+2xln(x)
If the resulting value of f'(x) is negative, the function is decreasing in that interval. If the result is positive, the function is increasing in that interval.
For the first interval (0, e^(-1/2)) , let x=0.2 be the test value.
f'(0.2)=0.2+2(0.2)ln(0.2)=-0.44 (Decreasing)
And for the second interval (e^(-1/2),oo) , let x=1 be the test value.
f'(1)=1+2(1)ln(1) = 1 (Increasing)
Thus, the function is decreasing at the interval (0, e^(-1/2)) . And it is increasing at (e^(-1/2),oo) .
(b) To determine the local maximum and local minimum, refer to the change of signs of f'(x) before and after the critical number.
If f'(x) changes from negative values to positive values, the local minimum of the function occurs at x=c. If f'(x) changes from positive values to negative values, the local maximum occurs at x=c.
For the critical number x=0, there is no change of signs of f'(x). Since negative numbers are not included in the domain.
And for the critical number x=e^(-1/2) , f'(x) changes from negative value to positive value. This means that the function has a local minimum at x=e^(-1/2) . The local minimum value of f(x) is:
f(x)=(x^2)ln(x)
f(e^(-1/2))=(e^(-1/2))^2ln (e^(-1/2))=-1/(2e)
Therefore, the local minimum is f(x)=-1/(2e) , which occurs at x=e^(-1/2) . And the function has no local maximum.
(c) To solve for the intervals in which the function is concave upward or downward, take the second derivative of f(x).
f'(x)=2xln(x) + x
f''(x) = 2x *1/x + 2*ln(x) + 1
f''(x)=2+2ln(x) + 1
f''(x) = 2lnx + 3
Set the second derivative equal to zero.
0=2ln(x) + 3
-3=2lnx
-3/2=ln x
x=e^(-3/2)
So the inflection occurs at x=e^(-3/2) . This means that at intervals (0,e^(-3/2)) and (e^(-3/2),oo) , the function has different concavities.
To determine its concavity, assign a test value for each interval. And plug-in them to the second derivative.
f''(x) = 2ln(x) + 3
If the resulting value of f"(x) is negative, the function is concave downward in that interval. If f"(x) is positive, the function is concave upward.
For the first interval (0,e^(-3/2)) , let the test value be x=0.1
f''(0.1) = 2ln(0.1) + 3 =-1.6 (Downward)
And for the second interval (e^(-3/2),oo) , let the test value be x=1.
f''(1)=2ln(1)+3=3 (Upward)
Thus, at interval (0,e^(-3/2)) the function is concave upward. And at (e^(-3/2),oo) , the function is concave downward.
To determine the inflection point, plug-in x=e^(-3/2) to the original function.
f(x) = x^2ln(x)
f(e^(-3/2))=(e^(-3/2))ln(e^(-3/2))=-3/(2e^3)
Therefore, the inflection point is (e^(-3/2), -3/(2e^3)) .