Chapter 4 – Applications of Differentiation 4.1 Maximum and Minimum Values Ex: Identify the following on this graph (Figure 2): A) Local Maximum B) Local Minimum C) Absolute (Global) Maximum D) Absolute (Global ) Minimum

.• • ← Not local

Max

Global Max

Absolute Max

Local t GlobalMin

N

^'t

#omhdx

.~•>x

v

Ex: Look at the same graph, but on the closed interval [1, 3] Identify the following:

A) Local Maximum B) Local Minimum C) Absolute (Global) Maximum D) Absolute (Global ) Minimum Question: What is f’(x) at the local extrema? What is f’(x) at the absolute extrema?

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•

()e¥mmmiI¥ilocal extreme .⇒T.am#./soabImeYFFxEo

or

sometimes it's an

endpoint

Parabola :

y=ax2+bx+cy'= Zaxtb = 0

Zaxtb = 0

Zax = - b

* Isamuvertex

Extreme Value Theorem If f is continuous on a closed interval [a, b], then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some numbers c and d in [a, b]

Question: Why does the interval need to be closed?

Fermat’s Theorem If f has a local maximum or minimum at c, and if f’(c) exists, then f’(c) = 0.

Note that this theorem is limited. It only goes in one direction, and it requires that the derivative exists in order to use it. Ex: Let f(x) = x3. Find f’(0) and discuss what this means for Fermat’s Theorem. Ex: Let f(x) = |x| What is f’(0)? Does f(x) have a max or min there anyway? Critical Number: A critical number of a function f is a number c in the domain of f such that either f’(c) = 0 or f’(c) does not exist.

To avoid asymptote S

Also ,we want f (a) and f ( b ) to exist

f ' ( x ) = 3×2 f'

G) = O

Fermat : If maxlmin - f '( 4=0fH={I¥¥c0oyen'¥#n=

' f'

G) DIE

Ex: Find the absolute maximum and absolute minimum values of f on the given interval: f(x) = 5 + 54x – 2x3 a) Find the values of f at the critical numbers of f in (a, b) b) Find the values of f at the endpoints of the interval c) The largest of the values from the first 2 steps is the absolute maximum value: The smallest of these values is the absolute minimum value:

on El ,u] possible local

fhu ) extreme

a tf 'Cx)= 54-6×2=0

54=6×2 (g) 5103→tF"" 3

9=× '

±3=X

f '

( x ) exists .

[ 1,11]ffD= -47

flu )= -2063.

113 at # 3

-2063 at ×= 11

4.2 The Mean Value Theorem

Rolle’s Theorem Let f be a function that satisfies all three of the following hypotheses: 1) f is continuous on the closed interval [a,b] 2) f is differentiable on the open interval (a,b) 3) f(a) = f(b) Then there is a number c in (a,b) such that f’(c) = 0.

Translation: If you throw a ball up into the air and catch it again, there will be some point in the middle when the velocity is 0.

The Mean Value Theorem Let f be a function that satisfies the following hypotheses: 1) f is continuous on the closed interval [a,b] 2) f is differentiable on the open interval (a,b) Then there is a number c in (a,b) such that f’(c) = ! " #!(%)

"#% or, equivalently, f(b) – f(a) = f’(c) (b – a)

Ex: Suppose that f(0) = -3 and f’(x) < 5 for all values of x. How large can f(2) possibly be? Hint: Look at f(b) – f(a) = f’(c) (b – a) over [0,2] Theorem: If f’(x) = 0 for all x in an interval (a,b), then f is constant on (a,b). Ex: Explain this using The Mean Value Theorem. Corollary: If f’(x) = g’(x) for all x in an interval (a,b), then f – g is a constant on (a,b). f(x) = g(x) + c, where c is constant.

4.3 How Derivatives Affect the Shape of a Graph Shapes, in general: f’(x) > 0 f’(x) < 0 f”(x) > 0 f”(x) < 0

First Derivative Test: Suppose c is a critical number of a continuous function f (f’(c) = 0 or f’ is not defined at c). a) If f’ changes from positive to negative at c, then f has a local maximum at c. b) If f’ changes from negative to positive at c, then f has a local minimum at c. c) If f’ does not change sign at c, then f has no local maximum or minimum at c.

Second Derivative Test: Suppose f’’ is continuous near c. a) If f’(c) = 0 and f”(c) > 0, then f has a local minimum at c. b) If f’(c) = 0 and f”(c) < 0, then f has a local maximum at c.

Ex: Sketch the graph f(x) = 36x + 3x2 – 2x3 using derivatives (If time) Example 7 on Page 296

4.4 Indeterminate Forms and L’Hospital’s Rule Ex: Calculate lim

*→,-.*/

Argh. Since lim

*→,0* = ∞ and lim

*→,45 = ∞,

this is called an indeterminate form of type ,,. Normally, we’d try to divide by x or look at the graph, since we weren’t able to evaluate this limit… until now. *Cue Exciting Music *

L’Hospital’s Rule Suppose f and g are differentiable and g’(x) ≠ 0 on an open interval l that contains a (except possibly at a.) Suppose that

lim*→%

6 4 = 089: lim*→%

; 4 = 0 or that

lim*→%

6 4 = ±∞89: lim*→%

; 4 = ±∞ (In other words, we have an indeterminate form of type == or ,,.) Then

lim*→%

6(4);(4) = lim*→%6′(4);′(4)

if the limit on the right side exists (or is ∞?@ − ∞. )

So… lim*→,

-.*/ = lim

*→,-.5* = lim*→,

-.5 = ∞

Ex: Find lim

*→CDE **#C

Big Warning – L’Hopital doesn’t always work: lim*→FG

HIJ*C#KLH* = ?

L’Hopital would suggest this: BUT, while sin π = 0, 1 – cos π = 1 – (-1) = 2, so the limit equals =5 = 0, not ==, so you can’t use L’Hopital. Ex: lim

*→=M4N94

(Hint: x ln x = DE *O

. )

4.5 Summary of Curve Sketching Guidelines for Sketching a Curve: 1) Domain 2) Intercepts 3) Symmetry If f(-x) = f(x) for all x, then f is even and the curve is symmetric about the y-axis If f(-x) = - f(x) for all x, then f is odd and the curve is symmetric about the origin 4) Asymptotes Vertical Asymptote: x = a is a vertical asymptote if any of the following is true: lim

*→%M6 4 = ∞ lim

*→%G6 4 = ∞

lim

*→%M6 4 = −∞ lim

*→%G6 4 = −∞

Horizontal Asymptotes: y = b is a horizontal asymptote if lim

*→,6 4 = P or lim

*→#,6 4 = P

Slant Asymptotes: y = mx + b is a slant asymptote if lim

*→,[ 6 4 − (R4 + P)] = 0

Ex: *U

*/VC = ? (Use Long Division) So f(x) = x - *

*/VC → x as x→ ±∞, which means that y = x is a slant asymptote. (Graph on Desmos) 5) Information from f’ and f’’ Intervals of Increase or Decrease Local Extrema Concavity and Points of Inflection 6) Additional Points as necessary

4.6 (Skip – Graphing Calculators) 4.7 Optimization Problems Ex: A cylindrical can is to be made to hold 1 Liter of oil. Find the dimensions that will minimize the cost of the metal to manufacture the can. (Hint: Minimize the surface area of the can.)

4.8 Newton’s Method How can we use calculus to estimate solutions that are hard to solve for? Newton’s Method uses f’(x) to improve a series of approximations that start with a single guess. Let’s say y = f(x) is a function whose x-intercepts are hard to find algebraically. We’re interested in a particular x-intercept x = r, and we’re going to use x1 as our starting guess for r. Step 1: Check if f(x1) = 0. If so, yay!, you’re done. If not, then we want to find another x – we’ll call it x2 – that’s closer to the true intercept r. Step 2: Find the derivative of f(x) at x1 and use it to find the tangent line (L) at x1.

Notice in the picture that the derivative helps us move closer to the actual intercept (r) when we look at the x-intercept (x2) of the tangent line (L). We’d love to use x2 as our next guess/approximation, but we have to find it first. We’ll use the formula for slope to find it: The slope of the tangent = W/#WO*/#*O

= !(*/)#!(*O)*/#*O

which leads us to

f’(x1) = !(*/)#!(*O)*/#*O

Since we’re looking for an x-intercept, we know that f(x2) = 0 f’(x1) = =#!(*O)*/#*O

Multiplying both sides by (x2 – x1) (x2 – x1) f’(x1) = 0 – f(x1) Solving for x2 we get x2 = x1 -

!(*O)!X(*O)

If f(x2) = 0, then we’re done. If not, then we can keep refining our guess with as many iterations as we want, using a generalized formula: xn+1 = xn -

!(*Y)!X(*Y)

Ex: Starting with x1 = 2, find the third approximation x3 to the root of the equation x3 – 2x – 5 = 0 f(x) = x3 – 2x – 5 f’(x) = 3x2 – 2 x2 = x1 -

*OU#5*O#Z[*O/#5

x2 = 2 - 5U#5∗5#Z[∗5/#5 = 2.1

x3 = x2 -

*/U#5*/#Z[*//#5

x3 = 2.1 - 5.C

U#5∗5.C#Z[∗5.C/#5 = 2.094568121…

When we decide to stop calculating successive values for x, how do we know how close we are to r, the actual root? Let’s look at x4: x4 = (2.0945…) - (5.=]^Z… )

U#5∗(5.=]^Z… )#Z[∗(5.=]^Z… )/#5 = 2.09455148…

Because we see that x3 and x4 match through 4 decimal places (2.094…), our approximation is correct to 4 decimal places.

4.9 Antiderivatives Ex: If f(x) = 3x + 2, find f’(x). Ex: If f’(x) = 3, find f(x)

Theorem: If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is F(x) + C where C is an arbitrary constant.

The book uses the phrase “the most general antiderivative” to indicate that you should use C.

Table of Antidifferentiation Formulas

Function a f(x) f(x) + g(x) xn (n ≠ -1) 14 ex cos x sin x sec2 x sec x tan x

11 − 45

1

1 +45 cosh x sinh x

Antiderivative a F(x) + C F(x) + G(x) + C *YMOJVC + C ln |x| + C ex + C sin x + C - cos x + C tan x + C sec x + C sin-1 x + C tan -1 x + C sinh x + C cosh x + C

Ex: Find all functions g such that g’(x) = 4 sin x + 5*

a# **

(Hint: Break g’(x) into individual rational expressions) Ex: Find f is f”(x) = 12x2 + 6x - 4, f(0) = 4and f(1) = 1.

Note: Recall that, if s(t) represents the position, v(t) = s’(t) gives the velocity and a(t) = s”(t) gives the acceleration.