Topic 5 Review (Ch. 4)

Critical Points

100

Find all critical points and identify them as local maximum points, local minimum points, or neither.

𝑦 = 2𝑥² - 2x + 16

Take the derivative:
y′=4x−2

Set derivative to 0:
4x−2=0

x= 1/2

Classify using second derivative:
 y'' = 4 > 0 → concave up → local minimum

Find y-value:
y=2(12)2−2(12)+16=15.5 
Local minimum at (1/2,15.5)

200

Find all critical points and identify them as local maximum points, local minimum points, or neither.

y = 4x^-2 + ln(x)

Take the derivative:
y′=4−1/x

Set derivative to 0:
4−1/x=0

x= 1/4

Classify using second derivative:
y′′=1/x^2>0 → concave up → local minimum

Find y-value:
y=4(14)−2+ln⁡(14)=1−2+ln⁡(0.25)≈−1−1.386=−2.386y 

Local minimum at (1/4,≈−2.386)

300

Find all critical points and identify them as local maximum points, local minimum points, or neither.

y = cos(2x) - x

Take the derivative:
y′=−2sin⁡(2x)−1

Set derivative to 0:
−2sin⁡(2x)−1=0

sin⁡(2x)=−1/2

Solve for x:

2x=7π/6,11π/6,⋯⇒x=7π/12,11π/12,…

Classify using second derivative:
y′′=−4cos⁡(2x)

• At x=7π/12x, cos⁡(2x)>0 ⇒ y′′<0 → local max

• At x=11π/12x, cos⁡(2x)<0 ⇒ y′′>0 → local min

Local max at 7π/12, local min at 11π/12 (and repeats every π)

400

Find all critical points and identify them as local maximum points, local minimum points, or neither.

f(x) = cot(x) + csc(x)

Take the derivative:

f′(x)=−csc⁡2(x)−csc⁡(x)cot⁡(x)

Set derivative to 0:

−csc⁡2(x)−csc⁡(x)cot⁡(x)=0

−csc⁡(x)(csc⁡(x)+cot⁡(x))=0

So,

csc(x)+cot(x)=0 

⇒

cot(x)=−csc(x)

use the identities

cos(x)=−1⇒x=π+2nπ, for n∈Z

Since f′′(x)>0 at x=π, it's a local minimum.

Local minimum at x=π+2nπ (for all integers n)

500

Find all critical points and identify them as local maximum points, local minimum points, or neither.

f(x) = sec(x) + 4x³ + x^-2

Take the derivative:

f′(x)=sec(x)tan(x)+12x^2 - 2/x^3

Set derivative to 0:

sec(x)tan(x) +12x^2 - 2/x^3 = 0

Test at x=1:

f'(1) = 12.89

not 0 so no crit point

use calc to find crit points.

use 2nd derivative to classify:

f′′(x)=sec(x)tan^2(x)+sec^3(x)+24x+6/x^4

Plug in x≈0.79 (from calc)

• All terms are positive → f′′(x)> 0

so , Local minimum at (0.79,5.15)

Local Extrema

100

Find the local extrema of the function:

f(x) = x³ - 3x² + 2

Step 1: First Derivative

f′(x)=3x²−6x

Step 2: Set Derivative = 0

3x²−6x=0 

3x(x−2)=0

x=0,2

Step 3: Second Derivative

f′′(x)=6x−6

• f′′(0)=−6 --> concave down --> local max at x=0
  
  

• f′′(2)=6 --> concave up --> local min at x=2

200

Find the local extrema of the function:

f(x) = x/ (x² +1)

Step 1: First Derivative (Quotient Rule)

f′(x)=(1-x²)/(x²+1)²

Step 2: Set Derivative = 0

1−x²=0 

x=±1  

Step 3: Second Derivative Test (Optional: First Derivative Test works too)

• Local maximum at x=−1, f(-1) = -1/2

• Local minimum at x=1, f(1) = 1/2

300

Find the local extrema of:

f(x)=x^2−4x+3

Local minimum at (2,−1)
There is no local maximum.

400

Find all local extrema of:

f(x)=x+1, x>0

Local minimum at (1,2)
No local maximum on x>0, since the function increases as x→∞ and also as x→0+

500

Find all local extrema on the interval [0,2π] for:

f(x)=sin⁡(x)cos⁡(x)

• Local maxima at (π/4,1/2) and (5π/4,1/2)

• Local minima at (3π/4,−1/2) and (7π/4,−1/2)

Global Extrema

100

Find the global maximum and minimum of

f(x)=sin⁡(x)+cos⁡(x)

on the interval [0,2π]

take the derivative:

f′(x)=cos(x)−sin(x)

set to 0:

x = π/4, 5π/4

 Evaluate f(x) at critical points and endpoints:

• f(0)=sin⁡(0)+cos⁡(0)=0+1=1

• f(2π)=sin⁡(2π)+cos⁡(2π)=0+1=1

• f(π/4)=≈1.41
  
  

• f(5π/4)=≈−1.41

• Global maximum at x=π/4: f(x)= root 2

• Global minimum at x=5π/4x: f(x)= - root 2

200

Find the global maximum and minimum of

f(x)=sin⁡(x)−cos⁡(x)

on the interval [0,2π]

derivative: f′(x)=cos(x)+sin(x)

set to 0:

x= 3π/4, 7π/4 

Evaluate f(x) at critical points and endpoints:

• f(0)=sin⁡(0)−cos⁡(0)=0−1=−1

• f(2π)=sin⁡(2π)−cos⁡(2π)=0−1=−1

• f(3π/4)=≈1.41

• f(7π/4)=≈−1.41

Answer:

• Global maximum at x=3π/4 = root 2

• Global minimum at x=7π/4x = - root 2

300

Find the local extrema of the function on the interval [0,2π]:

f(x)=sin(x)+cos(x)

Step 1: First Derivative

f′(x)=cos⁡(x)−sin⁡(x)

Step 2: Set Derivative = 0

cos⁡(x)=sin⁡(x) 

tan(x) = 1

x = π/4, 5π/4

Step 3: Plug into f(x)

• f(π/4)=sin⁡(π/4)+cos⁡(π/4) = (sq root 2)/2 + (sq root 2)/2 = (sq root 2) --> local maximum

• f(5π/4)= (- sq root 2) --> local minimum

400

Find the global extrema of the function on the interval [0, 5]:
f(x)= xe^-x

Step 1: Domain

• This function is defined for all real numbers because:

  • x is defined everywhere.

  • e^-x is defined and continuous everywhere.

• Since both parts are continuous, and the product of continuous functions is continuous, → f(x) is continuous and differentiable for all real x
• Since f(x) is on a closed interval, there has to be a global max/min.

Step 2: take the derivative using the product rule

f′(x)=e^-x(1−x)

Step 3: Find critical points on [0, 5]

Set f′(x)=0

e^-x(1−x) = 0 --> x=1 (since e^-x does not equal 0 for any x)

So, the critical point in the interval is:

• x=1

Step 4: Evaluate end points and critical point

We check:

• f(0)=0⋅e⁰=0

• f(1)=1⋅e^-1=1/e ≈0.3679

• f(5)=5⋅e^-5 ≈5⋅0.0067 =0.0337

Global maximum: at x=1, value = 1/e ≈0.3679

Global minimum: at x=0, value = 0 

500

Find the global maximum and minimum of

f(x)=ln⁡(x)/x

on the interval [1,4].

Step 1: Domain

The function f(x)=ln⁡(x)/x is:

• Defined only for x>0 because ln⁡(x) is undefined at x≤0

• On the interval [1,4], f(x) is continuous and differentiable, so the Extreme Value Theorem applies.

Step 2: Differentiate using the quotient rule

f'(x) = (1-ln(x))/x²

Step 3: find critical points

Set f'(x) = 0

1-ln(x) = 0

ln(x)= 1

x = e ≈ 2.718

Step 4:

We evaluate f(x)=ln⁡(x)/x at:

• f(1)=ln⁡(1)/1 = 0

• f(e)=ln⁡(e)/e = 1/e ≈ 0.3679f(e) 

• f(4)=ln⁡(4)/4 ≈ 1.3863/4 ≈ 0.3466

Global maximum: at x=e, value = 1/e ≈ 0.3679

Global minimum: at x=1, value = 0

Derivative Tests

100

For the function f(x)=x^2, use the first-derivative test to classify the critical point at x=0.

take derivative: 

f'(x) = 2x

2x= 0 --> x=0

First Derivative Test:

• Pick a point left of 0: x=−1
  f′(−1)=−2 → negative → decreasing

• Pick a point right of 0: x=1
  f′(1)=2→ positive → increasing

Since f′(x) changes from negative to positive,
→ Local minimum at x=0

local minimum at (0,0).

200

Find and classify the critical points of

f(x)=−x^2+4x

using the first derivative test.

take the derivative:

f'(x) = -2x +4

set to 0:

-2x +4 = 0

x=2

First Derivative Test:

• Left of 2 at x=1
  f′(1)=−2(1)+4=2>0→ increasing

• Right of 2 at x=3:
  f′(3)=−6+4=−2<0 → decreasing

So, f(x) changes from increasing to decreasing at x=2

local max at (2,4)

300

Find all local maximum and minimum points for 𝑓(𝑥)=sin𝑥 + cos𝑥 using the first derivative test.

take the derivative:

f'(x) = cosx - sinx

set derivative to 0:

x=π/4+nπ, for integers n

use 1st derivative test:

pick points around x=π/4 

at x=0 f'(x)= 1 --> positive 

at x= π/2, f'(x) = -1 --> negative

• Local maximum at x=π/4x
  f(π4)=sin⁡(π/4)+cos⁡(π/4)=root 2 

• Local minimum at x=5π/4
  f(5π4)=−root 2/2−root 2/2=− root 2

400

Use the first derivative test to find and classify all critical points of:

f(x)=x^3−3x

take the derivative: 

f'(x) = 3x^2 - 3

set to 0: 

0 = 3x^2 - 3

x= ± 1

 First Derivative Test:

• Interval (−∞,−1): Pick x=−2
  f′(−2)=3(4)−3=9>0 → increasing

• Interval (−1,1): Pick x=0
  f′(0)=−3<0 → decreasing

• Interval (1,∞): Pick x=2
  f′(2)=3(4)−3=9> 0 → increasing

• At x=−1, increasing → decreasing → local max

• At x=1, decreasing → increasing → local min

Find y-values:

• f(−1)=(−1)3−3(−1)=−1+3=2

• f(1)=13−3(1)=1−3=−2

Answer:

Local maximum at (−1,2)
Local minimum at (1,−2)

500

Let 𝑓(𝜃)=cos2(𝜃)−2sin(𝜃)

Find the intervals where 𝑓 is increasing and the intervals where 𝑓 is decreasing on [0,2𝜋][0,2π]. 

Use this information to classify the critical points of f as either local maximums, local minimums, or neither.

rewrite: (θ)=1−sin^2(θ)−2sin(θ)

take derivative and set to 0:

f′(θ)=−2cos(θ)(sin(θ)+1)=0

Solutions when:

• cos⁡(θ)=0⇒ θ=π/2,3π/2

• sin⁡(θ)=−1⇒θ=3π/2

So the critical points in [0,2π] are:

• θ=π/2

• θ=3π/2

use 1st derivative test and classify crit points: 

• Increasing on (π/2,3π/2)\

• Decreasing on [0,π/2)∪(3π/2,2π]

• Local minimum at θ=π/2

• Local maximum at θ=3π/2

Optimization/ Related Rates

100

A farmer has 100 meters of fencing to enclose a rectangular field.
What are the dimensions that maximize the area?

25 m by 25 m (a square)

200

A 10-ft ladder is leaning against a wall.
If the bottom slides away from the wall at 1 ft/sec, how fast is the top sliding down when the bottom is 6 ft from the wall?

3/4 ft/sec downward when the bottom is 6 ft from the wall

300

Find the dimensions of a cylinder with volume maximized, given the surface area is fixed at 100π100\pi100π square units.
Use the formula:

S=2πr^2+2πrh

V=πr^2h

Answer:
r=5
h=5
(Max volume when height = radius)

400

A balloon is rising at 5 ft/sec. A person is standing 20 feet away from the point directly below the balloon.
How fast is the distance between the person and the balloon increasing when the balloon is 15 feet high?

Answer:
Rate = 4 ft/sec

500

You’re making an open-top box from a 12×12 inch square sheet by cutting equal squares from each corner and folding up the sides.
What size square should you cut out to maximize the volume?

Answer:
x= 2