Answer to Question #222646 in Calculus for Caylin Adams

Question

Let f be the function:

f(x) = ln(2x)-(2x² +3), x > 0

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

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

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

(d) Find any inflection points

Answer 1

Let $f$ be the function defined as $f(x) = \ln(2x)-(2x^2 +3),\ x > 0.$

(a) The derivative of the function $f$ is $f'(x) = \frac{2}{2x}-4x=\frac{1-4x^2}{x}=\frac{(1-2x)(1+2x)}{x}.$ Since $x>0,$ the equality $f'(x)=0$ implies $x=\frac{1}{2}.$ If $x\in (0,\frac{1}{2})$ then $f'(x)=\frac{(1-2x)(1+2x)}{x}>0.$ If $x>\frac{1}{2},$ then $f'(x)<0.$ Therefore, on the interval $(0,\frac{1}{2})$ the function $f$ rises, on the interval $(\frac{1}{2},+\infty)$ the function $f$ falls.

(b) It follows from the previous item that $x=\frac{1}{2}$ is a point of maximum, $f(\frac{1}{2})=\ln 1-(\frac{1}{2}+3)=-\frac{7}{2}.$

(c) Let us find $f''(x)$:

$f''(x)=(\frac{1}{x}-4x)'=-\frac{1}{x^2}-4=-\frac{1+4x^2}{x^2}.$

It follows that $f''(x)<0$ for all $x>0,$ and hence on the interval $(0,+\infty)$ the graph of $f$ is concave down.

(d) It follows from the previous item that the function $f$ has no inflection point.