Visual Aids for Section 4.1, part 2 The First Derivative Testbarrus/mth141/sp20/Slides41b.pdf ·...

Visual Aids for Section 4.1, part 2The First Derivative Test

MTH 141

University of Rhode Island

MTH 141 (URI) Section 4.1 1 / 11

Our outcomes

Vocabulary: First Derivative Test for extrema.

Use the First Derivative Test to find and identify local maxima andminima of a given function.

Answer conceptual questions about the relationships among localextrema, critical points, and the signs of f ′(x) over intervals.

MTH 141 (URI) Section 4.1 2 / 11

How do you the local extrema of functions, given onlytheir formulas?

Our functions from the previous video, and a new one:

f (x) = x4 − 8x2 + 7

g(x) = arctan(x2)

h(x) = 4xe3x

j(x) = x4 − 6x2 − 8x

If f is defined on some interval (like a ≤ x ≤ b, for example), then anylocal maxima or local minima of f must occur at either critical pointsor at the interval endpoints.

Approach: We find local extrema in two steps:(1) Find the critical points by seeing where the first derivative is 0 or

undefined.(2) Decide whether each critical point is a local maximum, local

minimum, or neither. But how do we decide?

MTH 141 (URI) Section 4.1 3 / 11

How do you the local extrema of functions, given onlytheir formulas?

Our functions from the previous video, and a new one:

f (x) = x4 − 8x2 + 7

g(x) = arctan(x2)

h(x) = 4xe3x

j(x) = x4 − 6x2 − 8x

If f is defined on some interval (like a ≤ x ≤ b, for example), then anylocal maxima or local minima of f must occur at either critical pointsor at the interval endpoints.

Approach: We find local extrema in two steps:(1) Find the critical points by seeing where the first derivative is 0 or

undefined.(2) Decide whether each critical point is a local maximum, local

minimum, or neither. But how do we decide?

MTH 141 (URI) Section 4.1 3 / 11

How do you the local extrema of functions, given onlytheir formulas?

Our functions from the previous video, and a new one:

f (x) = x4 − 8x2 + 7

g(x) = arctan(x2)

h(x) = 4xe3x

j(x) = x4 − 6x2 − 8x

If f is defined on some interval (like a ≤ x ≤ b, for example), then anylocal maxima or local minima of f must occur at either critical pointsor at the interval endpoints.

Approach: We find local extrema in two steps:(1) Find the critical points by seeing where the first derivative is 0 or

undefined.(2) Decide whether each critical point is a local maximum, local

minimum, or neither. But how do we decide?

MTH 141 (URI) Section 4.1 3 / 11

How do you the local extrema of functions, given onlytheir formulas?

Our functions from the previous video, and a new one:

f (x) = x4 − 8x2 + 7

g(x) = arctan(x2)

h(x) = 4xe3x

j(x) = x4 − 6x2 − 8x

If f is defined on some interval (like a ≤ x ≤ b, for example), then anylocal maxima or local minima of f must occur at either critical pointsor at the interval endpoints.

Approach: We find local extrema in two steps:(1) Find the critical points by seeing where the first derivative is 0 or

undefined.(2) Decide whether each critical point is a local maximum, local

minimum, or neither.

But how do we decide?

MTH 141 (URI) Section 4.1 3 / 11

How do you the local extrema of functions, given onlytheir formulas?

Our functions from the previous video, and a new one:

f (x) = x4 − 8x2 + 7

g(x) = arctan(x2)

h(x) = 4xe3x

j(x) = x4 − 6x2 − 8x

If f is defined on some interval (like a ≤ x ≤ b, for example), then anylocal maxima or local minima of f must occur at either critical pointsor at the interval endpoints.

Approach: We find local extrema in two steps:(1) Find the critical points by seeing where the first derivative is 0 or

undefined.(2) Decide whether each critical point is a local maximum, local

minimum, or neither. But how do we decide?MTH 141 (URI) Section 4.1 3 / 11

One approach

The First-Derivative Test for Local Maxima and MinimaAfter finding the critical points of f , check the signs on f ′ around them. A

change from + to − means a local maximum, and a change from − to +

means a local minimum.

MTH 141 (URI) Section 4.1 4 / 11

One approach

The First-Derivative Test for Local Maxima and MinimaAfter finding the critical points of f , check the signs on f ′ around them. A

change from + to − means a local maximum, and a change from − to +

means a local minimum.

MTH 141 (URI) Section 4.1 4 / 11

Example: First Derivative Test

Find the local extrema of f (x) = x4 − 8x2 + 7.

From earlier: f ′(x) = 4x3 − 16x = 4x(x + 2)(x − 2)

Critical points at x = −2,0,2

MTH 141 (URI) Section 4.1 5 / 11

Example: First Derivative Test

Find the local extrema of f (x) = x4 − 8x2 + 7.

From earlier: f ′(x) = 4x3 − 16x = 4x(x + 2)(x − 2)

Critical points at x = −2,0,2

MTH 141 (URI) Section 4.1 5 / 11

Example: First Derivative Test

Find the local extrema of f (x) = x4 − 8x2 + 7.

From earlier: f ′(x) = 4x3 − 16x = 4x(x + 2)(x − 2)

Critical points at x = −2,0,2

MTH 141 (URI) Section 4.1 5 / 11

Example: First Derivative Test

Find the local extrema of f (x) = x4 − 8x2 + 7.

From earlier: f ′(x) = 4x3 − 16x = 4x(x + 2)(x − 2)

Critical points at x = −2,0,2

MTH 141 (URI) Section 4.1 5 / 11

Example: First Derivative Test

Find the local extrema of f (x) = x4 − 8x2 + 7.

From earlier: f ′(x) = 4x3 − 16x = 4x(x + 2)(x − 2)

Critical points at x = −2,0,2

MTH 141 (URI) Section 4.1 5 / 11

Example: First Derivative Test

Find the local extrema of f (x) = x4 − 8x2 + 7.

From earlier: f ′(x) = 4x3 − 16x = 4x(x + 2)(x − 2)

Critical points at x = −2,0,2

f ′(−3) = −12(−1)(−5)

MTH 141 (URI) Section 4.1 5 / 11

Example: First Derivative Test

Find the local extrema of f (x) = x4 − 8x2 + 7.

From earlier: f ′(x) = 4x3 − 16x = 4x(x + 2)(x − 2)

Critical points at x = −2,0,2

f ′(−1) = −4(1)(−3)

MTH 141 (URI) Section 4.1 5 / 11

Example: First Derivative Test

Find the local extrema of f (x) = x4 − 8x2 + 7.

From earlier: f ′(x) = 4x3 − 16x = 4x(x + 2)(x − 2)

Critical points at x = −2,0,2

f ′(1) = 4(3)(−1)

MTH 141 (URI) Section 4.1 5 / 11

Example: First Derivative Test

Find the local extrema of f (x) = x4 − 8x2 + 7.

From earlier: f ′(x) = 4x3 − 16x = 4x(x + 2)(x − 2)

Critical points at x = −2,0,2

f ′(1000000) = positive!

MTH 141 (URI) Section 4.1 5 / 11

Example: First Derivative Test

Find the local extrema of f (x) = x4 − 8x2 + 7.

From earlier: f ′(x) = 4x3 − 16x = 4x(x + 2)(x − 2)

Critical points at x = −2,0,2

f (x) has local minima at x = −2 and at x = 2, and a local maximum at x = 0.

MTH 141 (URI) Section 4.1 5 / 11

Example: First Derivative Test

f (x) = x4 − 8x2 + 7

MTH 141 (URI) Section 4.1 6 / 11

Another example

Find the local extrema of g(x) = arctan(x2).

From earlier: g′(x) =2x

1 + x4

Critical point at x = 0

MTH 141 (URI) Section 4.1 7 / 11

Another example

Find the local extrema of g(x) = arctan(x2).

From earlier: g′(x) =2x

1 + x4

Critical point at x = 0

MTH 141 (URI) Section 4.1 7 / 11

Another example

Find the local extrema of g(x) = arctan(x2).

From earlier: g′(x) =2x

1 + x4

Critical point at x = 0

MTH 141 (URI) Section 4.1 7 / 11

Another example

Find the local extrema of g(x) = arctan(x2).

From earlier: g′(x) =2x

1 + x4

Critical point at x = 0

MTH 141 (URI) Section 4.1 7 / 11

Another example

Find the local extrema of g(x) = arctan(x2).

From earlier: g′(x) =2x

1 + x4

Critical point at x = 0

MTH 141 (URI) Section 4.1 7 / 11

Another example

Find the local extrema of g(x) = arctan(x2).

From earlier: g′(x) =2x

1 + x4

Critical point at x = 0

g′(−1) = −1 and g′(1) = 1

MTH 141 (URI) Section 4.1 7 / 11

Another example

Find the local extrema of g(x) = arctan(x2).

From earlier: g′(x) =2x

1 + x4

Critical point at x = 0

g(x) has local minimum at x = 0.

MTH 141 (URI) Section 4.1 7 / 11

Another example

g(x) = arctan(x2)

MTH 141 (URI) Section 4.1 8 / 11

A final example

Find the local extrema of j(x) = x4 − 6x2 − 8x .

Derivative: j ′(x) = 4x3 − 12x − 8 = 4(x − 2)(x + 1)2

Critical points? x = −1 and x = 2

j ′(−1000000) = negative j ′(0) = −8 j ′(1000000) = positive

MTH 141 (URI) Section 4.1 9 / 11

A final example

Find the local extrema of j(x) = x4 − 6x2 − 8x .

Derivative: j ′(x) = 4x3 − 12x − 8 = 4(x − 2)(x + 1)2

Critical points?

x = −1 and x = 2

j ′(−1000000) = negative j ′(0) = −8 j ′(1000000) = positive

MTH 141 (URI) Section 4.1 9 / 11

A final example

Find the local extrema of j(x) = x4 − 6x2 − 8x .

Derivative: j ′(x) = 4x3 − 12x − 8 = 4(x − 2)(x + 1)2

Critical points? x = −1 and x = 2

j ′(−1000000) = negative j ′(0) = −8 j ′(1000000) = positive

MTH 141 (URI) Section 4.1 9 / 11

A final example

Find the local extrema of j(x) = x4 − 6x2 − 8x .

Derivative: j ′(x) = 4x3 − 12x − 8 = 4(x − 2)(x + 1)2

Critical points? x = −1 and x = 2

j ′(−1000000) = negative j ′(0) = −8 j ′(1000000) = positive

MTH 141 (URI) Section 4.1 9 / 11

A final example

Find the local extrema of j(x) = x4 − 6x2 − 8x .

Derivative: j ′(x) = 4x3 − 12x − 8 = 4(x − 2)(x + 1)2

Critical points? x = −1 and x = 2

j ′(−1000000) = negative j ′(0) = −8 j ′(1000000) = positive

MTH 141 (URI) Section 4.1 9 / 11

A final example

Find the local extrema of j(x) = x4 − 6x2 − 8x .

Derivative: j ′(x) = 4x3 − 12x − 8 = 4(x − 2)(x + 1)2

Critical points? x = −1 and x = 2

j ′(−1000000) = negative j ′(0) = −8 j ′(1000000) = positive

MTH 141 (URI) Section 4.1 9 / 11

A final example

Find the local extrema of j(x) = x4 − 6x2 − 8x .

Derivative: j ′(x) = 4x3 − 12x − 8 = 4(x − 2)(x + 1)2

Critical points? x = −1 and x = 2

j ′(−1000000) = negative j ′(0) = −8 j ′(1000000) = positive

j(x) has a local minimum at x = 2.

The point x = −1 is a critical point but not a local extremum.

MTH 141 (URI) Section 4.1 9 / 11

Another example

j(x) = x4 − 6x2 − 8x

MTH 141 (URI) Section 4.1 10 / 11

Our outcomes

Vocabulary: First Derivative Test for extrema.

Use the First Derivative Test to find and identify local maxima andminima of a given function.

Answer conceptual questions about the relationships among localextrema, critical points, and the signs of f ′(x) over intervals.

MTH 141 (URI) Section 4.1 11 / 11