Question

Let f be the function

f (x) = x2 − ln x,

where x > 1.

(a) Use the sign pattern for f'(x) to determine the intervals where f rises and where f

falls. (5)

(b) Determine the coordinates of the local extreme point(s). (2)

(c) Find f'' (x) and determine where the graph of f is concave up and where it is concave

down.

Answer 1

$$Solution$$

a. ) Given

$f(x)\ =\ x^2-ln\ x\\ f'(x)\ =\ 2x-\ \frac{1}{x}\\ f'(x)\ >\ 0\ on\ (1,\infin)$

hence f rises on interval $(1,\infin)$ and f does not fall on x>1

b. ) Determine the coordinates of the local extreme point(s).

x^2-ln\ x\ has no critical points, therefore there are no local extreme points.

(c) Find f'' (x) and determine where the graph of f is concave up and where it is concave

down.

$f(x)\ =\ x^2-ln\ x\\ f'(x)\ =\ 2x-\ \frac{1}{x},\\ f''(x)\ =\ 2+\ \frac{1}{x^2}\\ f''(x)\ >\ 0\ on\ (1,\infin)$

hence f is concave up at x>1 on the graph