Question #300344

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)


1
Expert's answer
2022-02-21T16:17:59-0500

(a)


f(x)=x2(2x+1)2f (x) =\dfrac{x^2 }{(-2x+1)^2}

2x+10=>x1/2-2x+1\not=0=>x\not=1/2

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


limx(1/2)x2(2x+1)2=\lim\limits_{x\to(1/2)^-}\dfrac{x^2 }{(-2x+1)^2} =\infin

limx(1/2)+x2(2x+1)2=\lim\limits_{x\to(1/2)^+}\dfrac{x^2 }{(-2x+1)^2} =\infin

Vertical asymptote: x=1/2x=1/2


limxx2(2x+1)2=limxx2/x2(2x/x+1/x)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}

=limx1(2+1/x)2=1/4=\lim\limits_{x\to-\infin}\dfrac{1}{(-2+1/x)^2}=1/4


limxx2(2x+1)2=limxx2/x2(2x/x+1/x)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}

=limx1(2+1/x)2=1/4=\lim\limits_{x\to\infin}\dfrac{1}{(-2+1/x)^2}=1/4

Horizontal asymptote: y=1/4y=1/4

There is no slant (oblique) asymptote.


(b)


f(x)=(x2(2x+1)2)f'(x)=(\dfrac{x^2 }{(-2x+1)^2} )'

=2x(2x+1)2x2(2)(2)(2x+1)(2x+1)4=\dfrac{2x(-2x+1)^2-x^2(2)(-2)(-2x+1) }{(-2x+1)^4}

=4x2+2x+4x2(2x+1)3=2x(2x+1)3=\dfrac{-4x^2+2x+4x^2}{(-2x+1)^3} =\dfrac{2x}{(-2x+1)^3}

Find critical number(s)


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

x=0x=0

(i)

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

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

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

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

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

(ii)

The function ff has a local minimum at x=0.x=0.

The function ff has no a local maximum.


(c)


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

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

=22x+1+6x(2x+1)4=8x+2(2x+1)4=2\cdot\dfrac{-2x+1+6x}{(-2x+1)^4} =\dfrac{8x+2}{(-2x+1)^4}

Find the inflection point(s)


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

x=1/4x=-1/4

(i)

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

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

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

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

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

(ii)

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


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