Single-Variable-Calculus · Unit 9 · Video 1 · Interactive Practice

Sketching Without Critical Points: (x+1)/(x+2)

IKey Formulas

Formula	Name	What it says
$f'(x) = \dfrac{(x+2) - (x+1)}{(x+2)^2} = \dfrac{1}{(x+2)^2}$	Quotient rule	Numerator $1$ is never zero: no critical points
$f(-2^{+}) = \dfrac{-1}{0^{+}} = -\infty,\quad f(-2^{-}) = \dfrac{-1}{0^{-}} = +\infty$	One-sided limits	$x = -2$ is a vertical asymptote
$f(x) = \dfrac{1 + \frac{1}{x}}{1 + \frac{2}{x}},\quad f(\pm\infty) = 1$	The ends	$y = 1$ is a horizontal asymptote
$f(x) = 1 - \dfrac{1}{x+2},\quad f''(x) = \dfrac{-2}{(x+2)^3}$	Hyperbola form	Concave up for $x \lt -2$ , concave down for $x \gt -2$

Key Insight: With no critical points the graph is drawn from its asymptotes. Since $f' \neq 0$ , neither branch can turn around, so each runs straight from one asymptote to the other.

IIVisualization 1 — The Four Edges of the Graph

Near $x = -2$ the denominator's sign decides between $\pm\infty$ ; far out, $\frac{1}{x}$ and $\frac{2}{x}$ vanish and $f \to 1$ .

approach x 0

IIIVisualization 2 — The Curve Is a Shifted Hyperbola

$f(x) = 1 - \frac{1}{x+2}$ is $\frac{1}{x}$ shifted left by $2$ , negated, and raised by $1$ .

step 1

IVVisualization 3 — Increasing on Each Piece, Not Across the Break

Every tangent slope $\frac{1}{(x+2)^2}$ is positive, yet $f(-3) = 2 \gt f(-1) = 0$ .

💡 $f' \gt 0$ on an interval makes $f$ increasing on that interval; $-\infty \lt x \lt \infty$ is not an interval of the domain, because $x = -2$ is missing.

VQuiz Questions

Problem 1 · The Derivative

Given: $f(x) = \dfrac{x+1}{x+2}$ — find $f'(x)$ with the quotient rule.

✅ Correct! The numerator simplifies to

1

, which is never zero, so

f

has no critical points.

❌ Close, but check the order. The quotient rule is (bottom)(top)′ minus (top)(bottom)′:

(x+2)\cdot 1 - (x+1)\cdot 1 = 1

, not

-1

❌ Not quite.

\frac{(x+1)'}{(x+2)'} = \frac{1}{1}

divides the derivatives, but the derivative of a quotient is not the quotient of the derivatives.

❌ Not quite.

2x + 3

comes from adding the two products. The quotient rule subtracts them:

(x+2) - (x+1) = 1

❌ Not quite. Apply

\left(\frac{u}{v}\right)' = \frac{v u' - u v'}{v^2}

with

u = x+1

v = x+2

Show solution

With $u = x + 1$ and $v = x + 2$ , both derivatives are $1$ :

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

Check with the hyperbola form $f(x) = 1 - (x+2)^{-1}$ : $f'(x) = 0 - \left(-(x+2)^{-2}\right) = \frac{1}{(x+2)^2}$ . The numerator $1$ is never zero, so there are no critical points.

Problem 2 · Where Is f Increasing?

Given: $f(x) = \dfrac{x+1}{x+2}$ has $f'(x) = \dfrac{1}{(x+2)^2} \gt 0$ for every $x \neq -2$ — find the correct statement.

✅ Correct! Increasing holds on each piece; across the break it fails, e.g.

f(-3) = 2

but

f(-1) = 0

❌ That is the shortcut the video warns against. Take

-3 \lt -1

f(-3) = 2

and

f(-1) = 0

, so a larger input gave a smaller output.

❌ Not quite. Critical points are where a graph could turn around. Having none means each branch keeps one direction, and

f' \gt 0

says that direction is up.

❌ Not quite.

f' \gt 0

on the left branch too: for instance

f(-5) = \frac{4}{3} \lt f(-3) = 2

❌ Not quite.

f' \gt 0

on each interval where

f

is defined; check what happens across

x = -2

Show solution

$f'(x) = \frac{1}{(x+2)^2}$ is $1$ over a square, positive wherever it is defined. On each of the intervals $-\infty \lt x \lt -2$ and $-2 \lt x \lt \infty$ , $f' \gt 0$ makes $f$ increasing.

The statement "increasing on $-\infty \lt x \lt \infty$ " is false: the left branch rises to $+\infty$ at $x = -2^{-}$ , and the right branch starts over at $-\infty$ . Concretely,

-3 \lt -1 \quad\text{but}\quad f(-3) = \frac{-2}{-1} = 2 \;\gt\; f(-1) = \frac{0}{1} = 0.

Problem 3 · Asymptotes of a New Function

Given: $g(x) = \dfrac{2x-3}{x-1}$ — find $g(1^{+})$ and the horizontal asymptote.

What is $g(1^{+})$ ?

What is the horizontal asymptote?

✅ Correct!

g(1^{+}) = \frac{-1}{0^{+}} = -\infty

and

g(\pm\infty) = \frac{2 - 3/x}{1 - 1/x} \to 2

: the same picture as the video, with asymptotes

x = 1

and

y = 2

❌ Check the sign. The numerator

2(1) - 3 = -1

is negative, and the denominator

1^{+} - 1 = 0^{+}

is positive: negative over tiny positive.

❌ Not quite.

2

is the value at the ends,

g(\pm\infty)

. Near

x = 1

the denominator goes to zero.

❌ Close, but…

-1

is only the numerator's limit. It is divided by a denominator shrinking to

0^{+}

❌ Not quite.

\frac{-3}{-1} = 3

g(0)

, the y-intercept. For the ends, divide top and bottom by

x

; the constants become

\frac{3}{x} \to 0

❌ Not quite.

y = 1

belongs to

\frac{x+1}{x+2}

. Here the leading coefficient on top is

2

❌ Check $g(1^{+})$ . Find the sign of the numerator and of the denominator separately as

x \to 1^{+}

❌ Check the asymptote. Divide numerator and denominator by

x

and let

x \to \pm\infty

Show solution

Vertical asymptote. The denominator vanishes at $x = 1$ . From the right:

g(1^{+}) = \frac{2(1) - 3}{1^{+} - 1} = \frac{-1}{0^{+}} = -\infty.

(From the left, $\frac{-1}{0^{-}} = +\infty$ .)

The ends. Divide numerator and denominator by $x$ :

g(x) = \frac{2 - \frac{3}{x}}{1 - \frac{1}{x}} \;\to\; \frac{2 - 0}{1 - 0} = 2 \quad\text{as } x \to \pm\infty.

So $y = 2$ is the horizontal asymptote. Equivalently $g(x) = 2 - \frac{1}{x-1}$ , the hyperbola $\frac{1}{x}$ shifted right by $1$ , negated, and raised by $2$ .

Problem 4 · Transfer: A Decreasing Version

Given: $h(x) = \dfrac{x+5}{x+3}$ — find $h(-3^{+})$ and describe where $h$ increases or decreases.

What is $h(-3^{+})$ ?

Which statement about $h$ is true?

✅ Correct!

h(x) = 1 + \frac{2}{x+3}

, so

h'(x) = -\frac{2}{(x+3)^2} \lt 0

: no critical points, each branch falls, and the right branch comes down from

+\infty

❌ Not quite — this is not the video's function. Here the numerator at

x = -3

-3 + 5 = 2

, positive: positive over

0^{+}

❌ Not quite.

1

h(\pm\infty)

, the horizontal asymptote, not the behaviour beside

x = -3

❌ Close, but…

2

is the numerator's limit; it is divided by

0^{+}

❌ Not quite. The quotient rule gives numerator

(x+3) - (x+5) = -2

, so

h' \lt 0

❌ Not across the break.

-4 \lt -2

but

h(-4) = -1 \lt h(-2) = 3

: monotonicity holds on each piece only.

❌ Not quite.

h(-5) = 0

makes

x = -5

an x-intercept. A critical point needs

h'(-5) = 0

, but

h'(-5) = -\frac{2}{4} \neq 0

❌ Check $h(-3^{+})$ . Find the signs of

x + 5

and

x + 3

x \to -3^{+}

❌ Check the second part. Compute

h'(x)

and remember that

x = -3

is not in the domain.

Show solution

From the right of $-3$ :

h(-3^{+}) = \frac{-3 + 5}{-3^{+} + 3} = \frac{2}{0^{+}} = +\infty.

Derivative.

h'(x) = \frac{(x+3)\cdot 1 - (x+5)\cdot 1}{(x+3)^2} = \frac{-2}{(x+3)^2} \lt 0 \quad (x \neq -3).

The numerator $-2$ is never zero: no critical points. So $h$ is decreasing on $-\infty \lt x \lt -3$ and on $-3 \lt x \lt \infty$ — but not across the break, since $h(-4) = -1 \lt h(-2) = 3$ .

The algebra trick confirms the picture: $h(x) = \frac{(x+3) + 2}{x+3} = 1 + \frac{2}{x+3}$ , a hyperbola with asymptotes $x = -3$ and $y = 1$ , the mirror image of the video's curve.

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