Question

Let f(x) = x^2 (ln x − 1).

(a) Find the critical numbers of f.

(b) Find the open interval(s) on which f is increasing and the open interval(s) on which f is decreasing.

(c) Find the local minimum value(s) and local maximum value(s) of f, if any.

(d) Find the open interval(s) where f is concave upward and the open interval(s) where f is concave downward.

(e) Find the inflection points of the graph of f, if any

Answer 1

 Step 1:

(a)

The function is  .

Apply derivative on each side with respect to x .

Product rule of derivatives : .

.

Derivative of the logarithmic function: .

Apply the power rule: .

Step 2:

To find the critical numbers of , equate to zero.

  

The critical points are . 

Solution:

The critical points are .

Answer 2

Step 1:

(b)

The critical points are .             (from (a))

Domain of the function:

Domain of the function is set of values of x which makes the function mathematically correct.

The logarithm function should be greater than zero.

Domain of the function is set of all values of x greater than 0.

Consider the test intervals.

Thus, The function is increasing on the interval and

The function is decreasing on the interval .

Step 2:

(c)

Consider .

Apply derivative on each side with respect to x .

Find :

.

If and , then has local minimum at .

By the above definition has local minimum at .

And never less than zero over the interval .

Therefore there are no local maximum to the function.

Solution:

(b)

The function is decreasing on the interval .

The function is increasing on the interval .

(c)

The function has local minimum at and there are no local maximum.

Answer 3

Step 1:

(d)

Consider .

Determination of concavity and inflection points : 

Equate to zero.

.

Thus, the inflection point is split the intervals into and .

Thus, the graph is concave up in the interval and 

The graph is concave down in the interval .

Step 2:

(e)

Inflection point is at :

Find the value of function at .

Substitute in the function.

.

Inflection point is  .

Solution:

(d)

The graph is concave up in the interval and 

The graph is concave down in the interval .

(e)

Inflection point is  .