Quiz 8 - Curve sketching

Problem 1

Question

Given the function \(y = 3x^4 - 4x^3\),

a)

Where is it increasing or decreasing?

b)

Where is it concave up or down?

c)

What are the local minima and maxima?

d)

Sketch the graph

Solution

a)

To find where it is increasing or decreasing, we can look at the sign of the derivative (the sign of the slope): \[ \begin{align} y &= 3x^4 - 4x^3 \\ y' &= 12x^3 - 12x^2 \\ &= 12x^2(x - 1) \end{align} \] This has critical points at \(x=0,1\). Since \(x^2\) is always positive, we can find our intervals by considering the sign of \((x-1)\) in the regions \((-\infty,0),(0,1),(1,\infty)\). Our intervals are then:

Increasing: \((1,\infty)\)
Decreasing: \((-\infty,0),(0,1)\)

b)

To find where it is concave up or down, we can look at the sign of the second derivative: \[ \begin{align} y' &= 12x^3 - 12x^2 \\ y'' &= 36x^2 - 24x \\ &= 12x(3x - 2) \end{align} \] This has inflection points at \(x=0,2/3\). Our intervals are then:

Concave down: \((0,2/3)\)
Concave up: \((-\infty,0),(2/3,\infty)\)

c)

Given the results in part a, we know that:

Local minima: \(x=1\)
Local maxima: None

d)

Problem 2

Question

Given the function \(y = 2 + 3x^2 - x^3\),

a)

Where is it increasing or decreasing?

b)

Where is it concave up or down?

c)

What are the local minima and maxima?

d)

Sketch the graph

Solution

a)

To find where it is increasing or decreasing, we can look at the sign of the derivative (the sign of the slope): \[ \begin{align} y &= 2 + 3x^2 - x^3\\ y' &= 6x - 3x^2\\ &= 3x(2 - x)\\ \end{align} \] This has critical points at \(x=0,2\). Our intervals are then:

Increasing: \((0,2)\)
Decreasing: \((-\infty,0),(2,\infty)\)

b)

To find where it is concave up or down, we can look at the sign of the second derivative: \[ \begin{align} y' &= 6x - 3x^2\\ y'' &= 6 - 6x \\ &= 6(1 - x) \end{align} \] This has an inflection point at \(x=1\). Our intervals are then:

Concave down: \((1,\infty)\)
Concave up: \((-\infty,1)\)

c)

Given the results in part a, we know that:

Local minima: \(x=0\)
Local maxima: \(x=2\)