Visual Aids for Section 4.1, part 1Derivatives and Critical Points

MTH 141

University of Rhode Island

MTH 141 (URI) Section 4.1 1 / 12

Our outcomes: Sec. 4.1, part 1

Vocabulary: local minimum of f at a point p, local maximum of fat p, local extrema, critical point of f , critical value of f .

Given a graph of a function f (x), find critical points, local maxima,and/or local minima visually.

Given a formula for f (x), find critical points using f ′(x).

Answer conceptual questions about the relationships among localextrema and critical points.

MTH 141 (URI) Section 4.1 2 / 12

Where is this function highest?

What can f tell you? What can f ′ tell you? What can f ′′ tell you?

f (x) = −2x3 − 18x2 − 46x − 30

y -coordinates (roots, end behavior)Examples: f (−4) = −6, f (−2) = 6, f (−1) = 0

f ′(x) = − 6x2 − 36x − 46slopes (increasing/decreasing)Examples: f ′(−3) = 8, f ′(0) = −46

f ′′(x) = − 12x − 36info about concavity (C.U., C.D.)Ex: f ′′(−4) = 12, f (−2) = −12, f ′′(0) = −36

MTH 141 (URI) Section 4.1 3 / 12

Where is this function highest?

What can f tell you? What can f ′ tell you? What can f ′′ tell you?

f (x) = −2x3 − 18x2 − 46x − 30

y -coordinates (roots, end behavior)Examples: f (−4) = −6, f (−2) = 6, f (−1) = 0

f ′(x) =

− 6x2 − 36x − 46slopes (increasing/decreasing)Examples: f ′(−3) = 8, f ′(0) = −46

f ′′(x) =

− 12x − 36info about concavity (C.U., C.D.)Ex: f ′′(−4) = 12, f (−2) = −12, f ′′(0) = −36

MTH 141 (URI) Section 4.1 3 / 12

Where is this function highest?

What can f tell you? What can f ′ tell you? What can f ′′ tell you?

f (x) = −2x3 − 18x2 − 46x − 30y -coordinates (roots, end behavior)Examples: f (−4) = −6, f (−2) = 6, f (−1) = 0

f ′(x) =

− 6x2 − 36x − 46slopes (increasing/decreasing)Examples: f ′(−3) = 8, f ′(0) = −46

f ′′(x) =

− 12x − 36info about concavity (C.U., C.D.)Ex: f ′′(−4) = 12, f (−2) = −12, f ′′(0) = −36

MTH 141 (URI) Section 4.1 3 / 12

Where is this function highest?

What can f tell you? What can f ′ tell you? What can f ′′ tell you?

f (x) = −2x3 − 18x2 − 46x − 30y -coordinates (roots, end behavior)Examples: f (−4) = −6, f (−2) = 6, f (−1) = 0

f ′(x) =

− 6x2 − 36x − 46slopes (increasing/decreasing)Examples: f ′(−3) = 8, f ′(0) = −46

f ′′(x) =

− 12x − 36info about concavity (C.U., C.D.)Ex: f ′′(−4) = 12, f (−2) = −12, f ′′(0) = −36

MTH 141 (URI) Section 4.1 3 / 12

Where is this function highest?

What can f tell you? What can f ′ tell you? What can f ′′ tell you?

f (x) = −2x3 − 18x2 − 46x − 30y -coordinates (roots, end behavior)Examples: f (−4) = −6, f (−2) = 6, f (−1) = 0

f ′(x) = − 6x2 − 36x − 46

slopes (increasing/decreasing)Examples: f ′(−3) = 8, f ′(0) = −46

f ′′(x) = − 12x − 36

info about concavity (C.U., C.D.)Ex: f ′′(−4) = 12, f (−2) = −12, f ′′(0) = −36

MTH 141 (URI) Section 4.1 3 / 12

Where is this function highest?

What can f tell you? What can f ′ tell you? What can f ′′ tell you?

f (x) = −2x3 − 18x2 − 46x − 30y -coordinates (roots, end behavior)Examples: f (−4) = −6, f (−2) = 6, f (−1) = 0

f ′(x) = − 6x2 − 36x − 46slopes (increasing/decreasing)Examples: f ′(−3) = 8, f ′(0) = −46

f ′′(x) = − 12x − 36

info about concavity (C.U., C.D.)Ex: f ′′(−4) = 12, f (−2) = −12, f ′′(0) = −36

MTH 141 (URI) Section 4.1 3 / 12

Where is this function highest?

What can f tell you? What can f ′ tell you? What can f ′′ tell you?

f (x) = −2x3 − 18x2 − 46x − 30y -coordinates (roots, end behavior)Examples: f (−4) = −6, f (−2) = 6, f (−1) = 0

f ′(x) = − 6x2 − 36x − 46slopes (increasing/decreasing)Examples: f ′(−3) = 8, f ′(0) = −46

f ′′(x) = − 12x − 36info about concavity (C.U., C.D.)Ex: f ′′(−4) = 12, f (−2) = −12, f ′′(0) = −36

MTH 141 (URI) Section 4.1 3 / 12

Where is this function highest?

What can f tell you? What can f ′ tell you? What can f ′′ tell you?

f (x) = −2x3 − 18x2 − 46x − 30y -coordinates (roots, end behavior)Examples: f (−4) = −6, f (−2) = 6, f (−1) = 0

f ′(x) = − 6x2 − 36x − 46slopes (increasing/decreasing)Examples: f ′(−3) = 8, f ′(0) = −46

f ′′(x) = − 12x − 36info about concavity (C.U., C.D.)Ex: f ′′(−4) = 12, f (−2) = −12, f ′′(0) = −36

MTH 141 (URI) Section 4.1 3 / 12

Where is this function highest?

What can f tell you? What can f ′ tell you? What can f ′′ tell you?

f (x) = −2x3 − 18x2 − 46x − 30y -coordinates (roots, end behavior)Examples: f (−4) = −6, f (−2) = 6, f (−1) = 0

f ′(x) = − 6x2 − 36x − 46slopes (increasing/decreasing)Examples: f ′(−3) = 8, f ′(0) = −46

f ′′(x) = − 12x − 36info about concavity (C.U., C.D.)Ex: f ′′(−4) = 12, f (−2) = −12, f ′′(0) = −36

MTH 141 (URI) Section 4.1 3 / 12

Vocabulary

f has a local minimum at p if f (p) is lower than (or equal to) everyvalue of f corresponding to inputs near p.

f has a local maximum at p if f (p) is higher than (or equal to) everyvalue of f corresponding to inputs near p.

Note: The plurals of minimum and maximum are minima and maxima. Anyvalue that is either a minimum or a maximum is called an extremum (plural:extrema).

A critical point of f is a point (in the domain, or on the graph) where f ′

is either 0 or undefined. The y -coordinate at a critical point is called acritical value of f .

MTH 141 (URI) Section 4.1 4 / 12

Practice with the vocabulary

Where are the critical points and local minima/maxima?

f (x) = −2x3 − 18x2 − 46x − 30

It looks like there is one local minimum,around x = −4.

It looks like there is one local maximum,around x = −2.

To find the critical points, look for where f ′ is either 0 or undefined.We look where the tangent line would be horizontal (or where it would

be a vertical line or nonexistent).

Here that seems to happen at the local minimum and local maximum.

MTH 141 (URI) Section 4.1 5 / 12

Practice with the vocabulary

Where are the critical points and local minima/maxima?

f (x) = −2x3 − 18x2 − 46x − 30

It looks like there is one local minimum,around x = −4.

It looks like there is one local maximum,around x = −2.

To find the critical points, look for where f ′ is either 0 or undefined.We look where the tangent line would be horizontal (or where it would

be a vertical line or nonexistent).

Here that seems to happen at the local minimum and local maximum.

MTH 141 (URI) Section 4.1 5 / 12

Practice with the vocabulary

Where are the critical points and local minima/maxima?

f (x) = −2x3 − 18x2 − 46x − 30

It looks like there is one local minimum,around x = −4.

It looks like there is one local maximum,around x = −2.

To find the critical points, look for where f ′ is either 0 or undefined.We look where the tangent line would be horizontal (or where it would

be a vertical line or nonexistent).

Here that seems to happen at the local minimum and local maximum.

MTH 141 (URI) Section 4.1 5 / 12

Practice with the vocabulary

Where are the critical points and local minima/maxima?

f (x) = −2x3 − 18x2 − 46x − 30

It looks like there is one local minimum,around x = −4.

It looks like there is one local maximum,around x = −2.

To find the critical points, look for where f ′ is either 0 or undefined.We look where the tangent line would be horizontal (or where it would

be a vertical line or nonexistent).

Here that seems to happen at the local minimum and local maximum.

MTH 141 (URI) Section 4.1 5 / 12

Practice with the vocabulary

Identify any local minima or maxima on the graph below, and anycritical points.

We have local minima at x = 2 and x = 3.5. We have local maxima at x = 1,x = 4, and x = 6.

We have a critical point where the derivative equals 0 at x = 2 and x = 3.5.We have a critical point where the derivative is undefined at x = 1, x = 4,

and x = 6.

MTH 141 (URI) Section 4.1 6 / 12

Practice with the vocabulary

Identify any local minima or maxima on the graph below, and anycritical points.

We have local minima at x = 2 and x = 3.5. We have local maxima at x = 1,x = 4, and x = 6.

We have a critical point where the derivative equals 0 at x = 2 and x = 3.5.We have a critical point where the derivative is undefined at x = 1, x = 4,

and x = 6.

MTH 141 (URI) Section 4.1 6 / 12

Practice with the vocabulary

Identify any local minima or maxima on the graph below, and anycritical points.

We have local minima at x = 2 and x = 3.5. We have local maxima at x = 1,x = 4, and x = 6.

We have a critical point where the derivative equals 0 at x = 2 and x = 3.5.We have a critical point where the derivative is undefined at x = 1, x = 4,

and x = 6.

MTH 141 (URI) Section 4.1 6 / 12

Testing for local mins and maxes

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.

Where are the local extrema forf (x) = −2x3 − 18x2 − 46x − 30?

We set f ′(x) = −6x2 − 36x − 46 = 0.

The quadratic formula gives us x =−(−36)±

√(−36)2 − 4(−6)(−46)

2(−6).

The critical points turn out to be x ≈ −4.15 and x ≈ −1.85.

MTH 141 (URI) Section 4.1 7 / 12

Testing for local mins and maxes

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.

Where are the local extrema forf (x) = −2x3 − 18x2 − 46x − 30?

We set f ′(x) = −6x2 − 36x − 46 = 0.

The quadratic formula gives us x =−(−36)±

√(−36)2 − 4(−6)(−46)

2(−6).

The critical points turn out to be x ≈ −4.15 and x ≈ −1.85.

MTH 141 (URI) Section 4.1 7 / 12

Testing for local mins and maxes

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.

Where are the local extrema forf (x) = −2x3 − 18x2 − 46x − 30?

We set f ′(x) = −6x2 − 36x − 46 = 0.

The quadratic formula gives us x =−(−36)±

√(−36)2 − 4(−6)(−46)

2(−6).

The critical points turn out to be x ≈ −4.15 and x ≈ −1.85.

MTH 141 (URI) Section 4.1 7 / 12

Testing for local mins and maxes

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.

Where are the local extrema forf (x) = −2x3 − 18x2 − 46x − 30?

We set f ′(x) = −6x2 − 36x − 46 = 0.

The quadratic formula gives us x =−(−36)±

√(−36)2 − 4(−6)(−46)

2(−6).

The critical points turn out to be x ≈ −4.15 and x ≈ −1.85.

MTH 141 (URI) Section 4.1 7 / 12

Testing for local mins and maxes

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.

Where are the local extrema forf (x) = −2x3 − 18x2 − 46x − 30?

We set f ′(x) = −6x2 − 36x − 46 = 0.

The quadratic formula gives us x =−(−36)±

√(−36)2 − 4(−6)(−46)

2(−6).

The critical points turn out to be x ≈ −4.15 and x ≈ −1.85.

MTH 141 (URI) Section 4.1 7 / 12

Exercises

For each of the functions below, find the critical points.

(a) f (x) = x4 − 8x2 + 7

We find f ′(x) = 4x3 − 16x . Setting it equal to 0 to find the critical point, we get

4x3 − 16x = 0

and so 4x(x2 − 4) = 4x(x + 2)(x − 2) = 0.

We see that the critical points are x = −2, x = 0, and x = 2.

(b) g(x) = arctan(x2)

Setting g′(x) =1

1 + (x2)2· 2x = 0,

we find that the only critical point is x = 0.

(c) h(x) = 4xe3x

Setting h′(x) = 4e3x + 12xe3x = 0,we get 4e3x (1 + 3x) = 0.We find that the only critical point isx = −1/3.

MTH 141 (URI) Section 4.1 8 / 12

Exercises

For each of the functions below, find the critical points.

(a) f (x) = x4 − 8x2 + 7

We find f ′(x) = 4x3 − 16x .

Setting it equal to 0 to find the critical point, we get

4x3 − 16x = 0

and so 4x(x2 − 4) = 4x(x + 2)(x − 2) = 0.

We see that the critical points are x = −2, x = 0, and x = 2.

(b) g(x) = arctan(x2)

Setting g′(x) =1

1 + (x2)2· 2x = 0,

we find that the only critical point is x = 0.

(c) h(x) = 4xe3x

Setting h′(x) = 4e3x + 12xe3x = 0,we get 4e3x (1 + 3x) = 0.We find that the only critical point isx = −1/3.

MTH 141 (URI) Section 4.1 8 / 12

Exercises

For each of the functions below, find the critical points.

(a) f (x) = x4 − 8x2 + 7

We find f ′(x) = 4x3 − 16x . Setting it equal to 0 to find the critical point, we get

4x3 − 16x = 0

and so 4x(x2 − 4) = 4x(x + 2)(x − 2) = 0.

We see that the critical points are x = −2, x = 0, and x = 2.

(b) g(x) = arctan(x2)

Setting g′(x) =1

1 + (x2)2· 2x = 0,

we find that the only critical point is x = 0.

(c) h(x) = 4xe3x

Setting h′(x) = 4e3x + 12xe3x = 0,we get 4e3x (1 + 3x) = 0.We find that the only critical point isx = −1/3.

MTH 141 (URI) Section 4.1 8 / 12

Exercises

For each of the functions below, find the critical points.

(a) f (x) = x4 − 8x2 + 7

We find f ′(x) = 4x3 − 16x . Setting it equal to 0 to find the critical point, we get

4x3 − 16x = 0 and so 4x(x2 − 4) = 4x(x + 2)(x − 2) = 0.

We see that the critical points are x = −2, x = 0, and x = 2.

(b) g(x) = arctan(x2)

Setting g′(x) =1

1 + (x2)2· 2x = 0,

we find that the only critical point is x = 0.

(c) h(x) = 4xe3x

Setting h′(x) = 4e3x + 12xe3x = 0,we get 4e3x (1 + 3x) = 0.We find that the only critical point isx = −1/3.

MTH 141 (URI) Section 4.1 8 / 12

Exercises

For each of the functions below, find the critical points.

(a) f (x) = x4 − 8x2 + 7

We find f ′(x) = 4x3 − 16x . Setting it equal to 0 to find the critical point, we get

4x3 − 16x = 0 and so 4x(x2 − 4) = 4x(x + 2)(x − 2) = 0.

We see that the critical points are x = −2, x = 0, and x = 2.

(b) g(x) = arctan(x2)

Setting g′(x) =1

1 + (x2)2· 2x = 0,

we find that the only critical point is x = 0.

(c) h(x) = 4xe3x

Setting h′(x) = 4e3x + 12xe3x = 0,we get 4e3x (1 + 3x) = 0.We find that the only critical point isx = −1/3.

MTH 141 (URI) Section 4.1 8 / 12

Exercises

For each of the functions below, find the critical points.

(a) f (x) = x4 − 8x2 + 7

We find f ′(x) = 4x3 − 16x . Setting it equal to 0 to find the critical point, we get

4x3 − 16x = 0 and so 4x(x2 − 4) = 4x(x + 2)(x − 2) = 0.

We see that the critical points are x = −2, x = 0, and x = 2.

(b) g(x) = arctan(x2)

Setting g′(x) =1

1 + (x2)2· 2x = 0,

we find that the only critical point is x = 0.

(c) h(x) = 4xe3x

Setting h′(x) = 4e3x + 12xe3x = 0,we get 4e3x (1 + 3x) = 0.We find that the only critical point isx = −1/3.

MTH 141 (URI) Section 4.1 8 / 12

Exercises

For each of the functions below, find the critical points.

(a) f (x) = x4 − 8x2 + 7

We find f ′(x) = 4x3 − 16x . Setting it equal to 0 to find the critical point, we get

4x3 − 16x = 0 and so 4x(x2 − 4) = 4x(x + 2)(x − 2) = 0.

We see that the critical points are x = −2, x = 0, and x = 2.

(b) g(x) = arctan(x2)

Setting g′(x) =1

1 + (x2)2· 2x = 0,

we find that the only critical point is x = 0.

(c) h(x) = 4xe3x

Setting h′(x) = 4e3x + 12xe3x = 0,we get 4e3x (1 + 3x) = 0.We find that the only critical point isx = −1/3.

MTH 141 (URI) Section 4.1 8 / 12

Exercises

For each of the functions below, find the critical points.

(a) f (x) = x4 − 8x2 + 7

We find f ′(x) = 4x3 − 16x . Setting it equal to 0 to find the critical point, we get

4x3 − 16x = 0 and so 4x(x2 − 4) = 4x(x + 2)(x − 2) = 0.

We see that the critical points are x = −2, x = 0, and x = 2.

(b) g(x) = arctan(x2)

Setting g′(x) =1

1 + (x2)2· 2x = 0,

we find that the only critical point is x = 0.

(c) h(x) = 4xe3x

Setting h′(x) = 4e3x + 12xe3x = 0,

we get 4e3x (1 + 3x) = 0.We find that the only critical point isx = −1/3.

MTH 141 (URI) Section 4.1 8 / 12

Exercises

For each of the functions below, find the critical points.

(a) f (x) = x4 − 8x2 + 7

We find f ′(x) = 4x3 − 16x . Setting it equal to 0 to find the critical point, we get

4x3 − 16x = 0 and so 4x(x2 − 4) = 4x(x + 2)(x − 2) = 0.

We see that the critical points are x = −2, x = 0, and x = 2.

(b) g(x) = arctan(x2)

Setting g′(x) =1

1 + (x2)2· 2x = 0,

we find that the only critical point is x = 0.

(c) h(x) = 4xe3x

Setting h′(x) = 4e3x + 12xe3x = 0,we get 4e3x (1 + 3x) = 0.

We find that the only critical point isx = −1/3.

MTH 141 (URI) Section 4.1 8 / 12

Exercises

For each of the functions below, find the critical points.

(a) f (x) = x4 − 8x2 + 7

We find f ′(x) = 4x3 − 16x . Setting it equal to 0 to find the critical point, we get

4x3 − 16x = 0 and so 4x(x2 − 4) = 4x(x + 2)(x − 2) = 0.

We see that the critical points are x = −2, x = 0, and x = 2.

(b) g(x) = arctan(x2)

Setting g′(x) =1

1 + (x2)2· 2x = 0,

we find that the only critical point is x = 0.

(c) h(x) = 4xe3x

Setting h′(x) = 4e3x + 12xe3x = 0,we get 4e3x (1 + 3x) = 0.We find that the only critical point isx = −1/3.

MTH 141 (URI) Section 4.1 8 / 12

Looking at the graphs

Remember, any local maxima orlocal minima of f must occur ateither critical points or at theinterval endpoints.

(a) f (x) = x4 − 8x2 + 7Critical points: x = −2,0,2

(b) g(x) = arctan(x2)

Critical points: x = 0(c) h(x) = 4xe3x

Critical points: x = −1/3

MTH 141 (URI) Section 4.1 9 / 12

The concept: a story

Suppose you live on a big, smooth hill, and you and your neighbor arechatting one day. “I wonder where the absolute top of this hill is,” yourneighbor says. “Sometimes I think I’m at the top, but then I wonder ifI’m actually a step or two away.”

“Well, let’s find out,” you say. “Just let me grab something from mytoolchest.” You grab a single tool, head out with your neighbor, and areable to tell exactly where the top is. What tool did you use, and how didyou use it?

MTH 141 (URI) Section 4.1 10 / 12

The concept: a story

Suppose you live on a big, smooth hill, and you and your neighbor arechatting one day. “I wonder where the absolute top of this hill is,” yourneighbor says. “Sometimes I think I’m at the top, but then I wonder ifI’m actually a step or two away.”

“Well, let’s find out,” you say. “Just let me grab something from mytoolchest.” You grab a single tool, head out with your neighbor, and areable to tell exactly where the top is. What tool did you use, and how didyou use it?

MTH 141 (URI) Section 4.1 10 / 12

Bonus challengeAn application from the past

In an earlier algebra class you probablylearned a formula for finding the locationof the vertex of a parabolay = ax2 + bx + c.

Now, you and I both know you definitely haven’t forgotten that formula,but let’s pretend for a minute that you have. Find it again (at least itsx-coordinate) using the calculus you now know.

MTH 141 (URI) Section 4.1 11 / 12

Our outcomes: Sec. 4.1, part 1

Vocabulary: local minimum of f at a point p, local maximum of fat p, local extrema, critical point of f , critical value of f .

Given a graph of a function f (x), find critical points, local maxima,and/or local minima visually.

Given a formula for f (x), find critical points using f ′(x).

Answer conceptual questions about the relationships among localextrema and critical points.

MTH 141 (URI) Section 4.1 12 / 12