Chapter 3 — Curve Sketching

First Derivative & Critical Points

100

For f(x)=x³−6x²+9x+4,

find all critical numbers.

x=1, 3

200

For f(x)=x⁴−4x³, determine the intervals on which the function is increasing and decreasing.

Decreasing: (−∞,3)

Increasing: (3,∞) 

300

Find the local maximum and local minimum of

f(x)=x³−3x²−9x+5. 

Local maximum (−1,10) 

Local minimum (3,−22) 

400

A function has f′(x)=x²(x−2)(x+3).

Identify its critical numbers and determine whether the function changes from increasing to decreasing or decreasing to increasing at each critical number.

Critical numbers:

x=−3, 0, 2

Sign analysis gives:

• x=−3: increasing → decreasing → local maximum
• x=0: increasing → increasing → neither
• x=2: decreasing → increasing → local minimum 

500

Given f′(x)=(x−1)²(x+2), determine all local extrema of f(x).

Critical numbers:

x=−2, x=1

At x=−2, f′ changes from negative to positive:

Local minimum at x=−2

At x=1, f′ does not change sign:

No local extremum at x=1 

Second Derivative & Concavity

100

Find the intervals of concavity for

f(x)=x³−6x²+9x+1. 

f′′(x)=6x−12

Thus, Concave down on (−∞,2) Concave up on (2,∞) 

200

Find the point of inflection of

f(x)=x³−9x²+24x−5. 

Answer:

f′′(x)=6x−18

Set f′′(x)=0:

x=3, f(3)=27−81+72−5=13

Therefore: (3,13) 

300

 For f(x)=x⁴−4x³+10, find the intervals where the graph is concave up and concave down.

f′′(x)=12x²−24x=12x(x−2)

Concave up: (−∞,0)∪(2,∞)

Concave down: (0,2) 

400

Does f′′(x)=0 always indicate a point of inflection? Explain

Answer: No. The concavity must actually change at the point.

500

Determine all points of inflection of

f(x)=x⁵−5x⁴+5x³. 

Answer:

f′′(x)=20x3−60x2+30x =10x(2x2−6x+3)

Thus: x=0, x=(3^1/2−3)/2, x=(3^1/2+3)/2

All three produce changes in concavity, so these are the inflection x-values.

Curve Analysis

100

What information does the first derivative provide when sketching a curve?

It determines where the function is increasing, decreasing, and has critical points.

200

What information does the second derivative provide when sketching a curve?

It determines concavity and helps identify possible points of inflection.

300

For f(x)=x³−3x, determine the local maximum and local minimum.

f′(x)=3x²−3=3(x−1)(x+1)

Critical numbers: x=−1, 1 f(−1)=2,f(1)=−2

Therefore: Local maximum (−1,2) Local minimum (1,−2)

400

For f(x)=x⁴−2x², find all critical points and classify them.

f′(x)=4x³−4x=4x(x−1)(x+1)

Critical numbers: x=−1,0,1

Values: f(−1)=−1,f(0)=0,f(1)=−1

Therefore: Local minima (−1,−1),(1,−1) Local maximum (0,0) 

500

 Sketch the general shape of f(x)=x⁴−4x².

Identify its critical points, local extrema, and points of inflection.

Critical numbers:     x=−2^1/2, 0, 2^1/2

Values:             f(±2^1/2)=−4, f(0)=0

So: Local minima (±2^1/2,−4) Local maximum (0,0) 

Also, f′′(x)=12x²−8 

        f′′(x)=0⇒x=±32

The points of inflection are

(−(2/3)^1/2, −20/9),((2/3)^1/2,−20/9) 

Rational Functions & Asymptotes

100

Find the vertical asymptote of

f(x)=(x+2)/(x-3). 

x=3

200

Find the horizontal asymptote of

f(x)=(3x²−1)/(x²+4). 

y=3

300

Determine the vertical and horizontal asymptotes of

f(x)=(2x+5)/(x−4). 

x=4, y=2

400

For f(x)=(x²−1)/(x−1),

identify any discontinuity and determine whether it is a vertical asymptote.

There is a hole at (1,2)

There is no vertical asymptote

500

Determine the vertical and horizontal asymptotes of

f(x)=(2x³−5x²+x+3)/(x²−1). 

Vertical asymptote: x=−1

There is a hole at x=1.

Polynomial division gives: f(x)=(2x−5)+[2/(x+1)]

Therefore, the function has a slant asymptote:

y=2x−5 

Advanced Curve-Sketching Challenges

100

 A function has

f′(x)=−(x−2)²(x+1).

Without finding f(x), determine where the function is increasing and decreasing.

Increasing:   (−∞,−1)

Decreasing:  (−1,∞) 

200

A function satisfies f′(x)=x(x−2)(x+3).

At which critical number does the function have a local maximum?

Critical numbers: x=−3, 0, 2

The sign changes from + to − at x=0.

x=0 

300

Find the equation of the tangent line to

f(x)=x³−6x²+9x+2

at its local maximum.

f′(x)=3(x−1)(x−3)

Local maximum occurs at x=1.   f(1)=6

Since the tangent at a local maximum is horizontal: y=6 

400

Determine the number of local extrema of

f(x)=x⁵−5x³+4x. 

f′(x)=5x⁴−15x²+4

Let u=x²:        5u²−15u+4=0 

u=(15±145^1/2)/10 

Both values are positive, producing four distinct critical numbers: 

The derivative changes sign at each, so there are

4 local extrema 

500

Perform a complete curve analysis of

f(x)=x²/(x-1)

Domain: x²=1

Intercept: x²=0⇒(0,0)

Vertical asymptote:  x=1

Slant asymptote: y=x+1 

Critical numbers: x=0, 2

Increasing: (−∞,0)∪(2,∞)

Decreasing: (0,1)∪(1,2)

Function values: f(0)=0,f(2)=4

Therefore: Local maximum (0,0) Local minimum (2,4) 

Concave down: (−∞,1)

Concave up: (1,∞)

Concavity changes at x=1, but f(1) is undefined, so there is no actual point of inflection.