Question

1. given the following function:

$f(x) = \frac{-1}{2x + 1}$
determine
a) domain
b) behavior near the asymptotes
c) end behavior
d) $x$-intercept(s)/$y$-intercept/max or min value
e) sketch
f) range
g) when $f(x) > 0$

Explanation:

Step1: Find Domain (a)

The function is a rational function \( f(x)=\frac{-1}{2x + 1} \). The denominator cannot be zero. So, solve \( 2x+1
eq0 \). Subtract 1 from both sides: \( 2x
eq - 1 \), then divide by 2: \( x
eq-\frac{1}{2} \). So the domain is all real numbers except \( x = -\frac{1}{2} \), or in interval notation \( (-\infty,-\frac{1}{2})\cup(-\frac{1}{2},\infty) \).

Step2: Behavior near Asymptotes (b)

Vertical asymptote at \( x = -\frac{1}{2} \). For \( x\to-\frac{1}{2}^- \) (left of \( -\frac{1}{2} \)), \( 2x + 1\to0^- \), so \( f(x)=\frac{-1}{2x + 1}\to\frac{-1}{0^-}=+\infty \). For \( x\to-\frac{1}{2}^+ \) (right of \( -\frac{1}{2} \)), \( 2x + 1\to0^+ \), so \( f(x)=\frac{-1}{2x + 1}\to\frac{-1}{0^+}=-\infty \). Horizontal asymptote: The degree of numerator (0) and denominator (1). When degree of numerator < degree of denominator, horizontal asymptote is \( y = 0 \). As \( x\to\pm\infty \), \( f(x)\to0 \). For \( x\to+\infty \), \( 2x + 1\to+\infty \), so \( f(x)=\frac{-1}{+\infty}\to0^- \). For \( x\to-\infty \), \( 2x + 1\to-\infty \), so \( f(x)=\frac{-1}{-\infty}\to0^+ \).

Step3: End Behavior (c)

As \( x\to+\infty \), \( f(x)\to0 \) from below (negative side). As \( x\to-\infty \), \( f(x)\to0 \) from above (positive side).

Step4: Intercepts and Extrema (d)

• x - intercept: Set \( f(x)=0 \), \( \frac{-1}{2x + 1}=0 \). But \( - 1

eq0 \), so no x - intercept.

• y - intercept: Set \( x = 0 \), \( f(0)=\frac{-1}{2(0)+1}=-1 \). So y - intercept is \( (0,-1) \).
• Max or min: The function is a hyperbola (rational function), it has no maximum or minimum values (since it approaches \( \pm\infty \) near vertical asymptote and 0 at infinity).

Step5: Sketch (e)

Draw vertical asymptote \( x = -\frac{1}{2} \), horizontal asymptote \( y = 0 \). Plot the y - intercept \( (0,-1) \). For \( x>-\frac{1}{2} \), as \( x\to-\frac{1}{2}^+ \), \( f(x)\to-\infty \), and as \( x\to+\infty \), \( f(x)\to0^- \). For \( x<-\frac{1}{2} \), as \( x\to-\frac{1}{2}^- \), \( f(x)\to+\infty \), and as \( x\to-\infty \), \( f(x)\to0^+ \).

Step6: Range (f)

Let \( y=\frac{-1}{2x + 1} \). Solve for x: \( y(2x + 1)=-1 \), \( 2xy+y=-1 \), \( 2xy=-1 - y \), \( x=\frac{-1 - y}{2y} \). The denominator \( 2y
eq0 \), so \( y
eq0 \). So range is all real numbers except \( y = 0 \), or \( (-\infty,0)\cup(0,\infty) \).

Step7: Solve \( f(x)>0 \) (g)

\( \frac{-1}{2x + 1}>0 \). The numerator is - 1 (negative). For the fraction to be positive, the denominator must be negative (negative/negative = positive). So \( 2x+1<0 \), \( 2x<-1 \), \( x<-\frac{1}{2} \).

Answer:

s:
a) Domain: \( (-\infty,-\frac{1}{2})\cup(-\frac{1}{2},\infty) \)
b) Near \( x = -\frac{1}{2}^- \), \( f(x)\to+\infty \); near \( x = -\frac{1}{2}^+ \), \( f(x)\to-\infty \); near \( y = 0^- \) as \( x\to+\infty \), near \( y = 0^+ \) as \( x\to-\infty \)
c) As \( x\to+\infty \), \( f(x)\to0^- \); as \( x\to-\infty \), \( f(x)\to0^+ \)
d) x - intercept: None; y - intercept: \( (0, - 1) \); No max/min
e) Sketch: Vertical asymptote \( x=-\frac{1}{2} \), horizontal asymptote \( y = 0 \), passes through \( (0,-1) \), left of \( x = -\frac{1}{2} \) the graph is in the second quadrant (approaching \( +\infty \) as \( x\to-\frac{1}{2}^- \) and \( 0^+ \) as \( x\to-\infty \)), right of \( x = -\frac{1}{2} \) the graph is in the fourth quadrant (approaching \( -\infty \) as \( x\to-\frac{1}{2}^+ \) and \( 0^- \) as \( x\to+\infty \))
f) Range: \( (-\infty,0)\cup(0,\infty) \)
g) \( x<-\frac{1}{2} \)