Answer to Question #300344 in Calculus for Karly

Question

1. Let f be the function deÖned by

f (x) = x^2/(-2x + 1)^2

(a) Determine the vertical and horizontal asymptotes (show all limits).

(b) Use the sign pattern for f'(x) to determine

(i) the interval(s) over which f rises and where it falls;

(ii) the local extrema.

(c) Use the sign pattern for f''(x) to determine

(i) where the graph of f is concave up and where it is concave down.

(ii) the inflection points (if any)

Answer 1

(a)

$$f (x) =\dfrac{x^2 }{(-2x+1)^2}$$

$$-2x+1\not=0=>x\not=1/2$$

$D(f): (-\infin, 1/2)\cup (1/2 , \infin)$

$$\lim\limits_{x\to(1/2)^-}\dfrac{x^2 }{(-2x+1)^2} =\infin$$

$$\lim\limits_{x\to(1/2)^+}\dfrac{x^2 }{(-2x+1)^2} =\infin$$

Vertical asymptote: $x=1/2$

$$\lim\limits_{x\to-\infin}\dfrac{x^2 }{(-2x+1)^2} =\lim\limits_{x\to-\infin}\dfrac{x^2/x^2 }{(-2x/x+1/x)^2}$$

$$=\lim\limits_{x\to-\infin}\dfrac{1}{(-2+1/x)^2}=1/4$$

$$\lim\limits_{x\to\infin}\dfrac{x^2 }{(-2x+1)^2} =\lim\limits_{x\to\infin}\dfrac{x^2/x^2 }{(-2x/x+1/x)^2}$$

$$=\lim\limits_{x\to\infin}\dfrac{1}{(-2+1/x)^2}=1/4$$

Horizontal asymptote: $y=1/4$

There is no slant (oblique) asymptote.

(b)

$$f'(x)=(\dfrac{x^2 }{(-2x+1)^2} )'$$

$$=\dfrac{2x(-2x+1)^2-x^2(2)(-2)(-2x+1) }{(-2x+1)^4}$$

$$=\dfrac{-4x^2+2x+4x^2}{(-2x+1)^3} =\dfrac{2x}{(-2x+1)^3}$$

Find critical number(s)

$$f'(x)=0=>\dfrac{2x}{(-2x+1)^3} =0$$

$$x=0$$

(i)

If $x<0, f'(x)<0, f(x)$ decreases.

If $0<x<1/2, f'(x)>0, f(x)$ increases.

If $x>1/2, f'(x)<0, f(x)$ decreases.

The function $f$ increases on $(0, 1/2).$

The function $f$ decreases on $(-\infin, 0)\cup (1/2, \infin).$

(ii)

The function $f$ has a local minimum at $x=0.$

The function $f$ has no a local maximum.

(c)

$$f''(x)=(\dfrac{2x }{(-2x+1)^3} )'$$

$$=2\cdot\dfrac{(-2x+1)^3-x(3)(-2)(-2x+1)^2 }{(-2x+1)^6}$$

$$=2\cdot\dfrac{-2x+1+6x}{(-2x+1)^4} =\dfrac{8x+2}{(-2x+1)^4}$$

Find the inflection point(s)

$$f''(x)=0=>\dfrac{8x+2}{(-2x+1)^4} =0$$

$$x=-1/4$$

(i)

If $x<-1/4, f''(x)<0, f(x)$ is concave down.

If $-1/4<x<1/2, f''(x)>0, f(x)$ is concave up.

If $x>1/2, f''(x)>0, f(x)$ is concave up.

The function $f$ is concave up on $(-1/4, 1/2)\cup (1/2, \infin).$

The function $f$ is concave down on $(-\infin, -1/4).$

(ii)

The function $f$ has inflection point at $x=-1/4.$