Applications of differentiation

MATH 150 Spring 2012

Applications of differentiation

Related rates

Related rates problems involve finding a rate at which a quantity changesby relating that quantity to other quantities whose rates of change areknown (the rate of change is usually taken with respect to time).

Example

If y = x3 + 2x and dxdt

= 5, find dydt

when x = 2.

Example

If x2 + y2 = 25 and dy

dt= 6, find dx

dtwhen y = 4.

Applications of differentiation

Related rates

Example

Air is being pumped into a spherical balloon so that its volume increasesat a rate of 100cm3/s. How fast is the radius of the balloon increasingwhen its diameter is 50cm?

Example

A particle is moving along the curve y = 3√2x + 1. As the particle

passes through the point (4, 9), its x-coordinate increases at a rate of 2units per second. Find the rate of change of the distance from theparticle to the origin at this instant.

Applications of differentiation

Maximum and minimum values

Definition

1 A function f (x) has an absolute (or global) maximum at x = c iff (c) > f (x) for every x ∈ Dom(f ). f (c) is called the globalmaximum in this case.

2 A function f (x) has an absolute (or global) minimum at x = c iff (c) 6 f (x) for every x ∈ Dom(f ). f (c) is called the globalminimum in this case.

Extremum refers to both maximum and minimum.

Applications of differentiation

Maximum and minimum values

Example (positive example)

Find out if f (x) = −x2 + 1 has a global maximum or minimum.

Example (negative examples)

Consider

(a) f (x) = ex

(b) g(x) = tan−1 x

(c) y = x3

Applications of differentiation

Maximum and minimum values

Definition

1 A function f (x) has a local maximum at x = c if f (c) > f (x) forevery x near c. f (c) is called the local maximum in this case.

2 A function f (x) has a local minimum at x = c if f (c) 6 f (x) forevery x near c. f (c) is called the local minimum in this case.

Example

Consider y = x3 − 3x .

Applications of differentiation

Finding absolute extrema on a closed segment

Fact (Extreme Value Theorem)

If f (x) is continuous on a closed interval [a, b], then f (x) attains itsabsolute maximum and absolute minimum values on [a, b].

Observe that this does not apply to discontinuous functions.

Example

Consider

f (x) =

{

x , if 0 6 x < 1,3− x , if 1 6 x 6 2.

on [0, 2].

Applications of differentiation

Finding absolute extrema on a closed segment

The Extreme Value Theorem says that both absolute maximum andabsolute minimum values exist on closed interval for a continuousfunction. The question is how to find them.Since every absolute extremum is also a local extremum, one may try tolook for local extrema first.

Fact (Fermat’s Theorem)

If f (x) has a local maximum or minimum at c and if f ′(c) exists thenf ′(c) = 0.

Hence, the following definition.

Definition

A critical number of a function f (x) is a number c in the domain of f (x)such that either f ′(c) = 0, or f ′(c) does not exist.

Applications of differentiation

Finding absolute extrema on a closed segment

Fact

If f (x) has a local maximum or minimum at c then c is a critical numberof f (x).

So, based on the previous fact, here is the Closed Interval Method to findthe absolute maximum and minimum of a continuous function f (x) on aclosed interval [a, b]:

find the values of f (x) at the critical numbers of f (x) in (a, b),

find the values of f (x) at the endpoints of the interval,

the largest of the found values is the absolute maximum value, thesmallest is the absolute minimum value.

Example

(a) f (x) = x3 − 3x + 1 on [0, 3]

(b) f (x) = x2−4x2+4

on [−4, 4]

Applications of differentiation

The Mean Value Theorem

Fact (The Mean Value Theorem)

If f (x) is continuous on [a, b] and differentiable on (a, b) function, thenthere exists a number c between a and b such that

f ′(c) =f (b)− f (a)

b − a,

or, equivalently,f (b)− f (a) = f ′(c)(b − a).

Example

Suppose f (0) = −3 and f ′(x) 6 5 for every x . How large can f (2)possibly be?

Applications of differentiation

The Mean Value Theorem

The Mean Value Theorem has several interesting corollaries.

Fact (Rolle’s Theorem)

If f (x) is continuous on [a, b] and differentiable on (a, b) function, andf (a) = f (b) then there exists a number c ∈ (a, b) such that f ′(c) = 0.

Example

Prove that x3 + x − 1 = 0 has exactly one real root.

Applications of differentiation

The Mean Value Theorem

Fact

If f (x) is continuous on [a, b] and differentiable on (a, b) function, andf ′(c) = 0 for every c ∈ (a, b) then f is a constant function on (a, b).

Fact

If f (x) and g(x) are continuous on [a, b] and differentiable on (a, b)functions, and f ′(x) = g ′(x) for every x ∈ (a, b) then f − g is a constantfunction on (a, b).

Example

Prove that tan−1 x + cot−1 x = π

2.

Applications of differentiation

Curve sketching

First of all, recall the following fact.

Fact

If f ′(x) > 0 (f ′(x) < 0) on an interval then f (x) is increasing(decreasing) on that interval.

It can be proved using the Mean Value Theorem.Next, we know how to find local maxima and minima of a given function.Namely, the following method is called the First Derivative Test to checkif a critical number c of a continuous function f (x) gives a localextremum:

if f ′(x) changes from positive to negative at c , then f (x) has a localmaximum at c ,

if f ′(x) changes from negative to positive at c , then f (x) has a localminimum at c ,

if f ′(c) does not change its sign at c , then f (x) has no extremum atc .

Applications of differentiation

Curve sketching

Example

Find where the function f (x) = 3x4 − 4x3 − 12x2 + 5 is increasing andwhere it is decreasing. Find all local extrema of f (x).

Recall that the graph of a function f (x) is concave upward (downward)on an interval if it is above (below) the tangent line at every point ofthat interval.

Fact

If f ′′(x) > 0 (f ′′(x) < 0) on an interval then the graph of f (x) is concaveupward (downward) on that interval.

Applications of differentiation

Curve sketching

Now, using all these facts and methods we can sketch the graph of agiven function.

Example

Sketch the graph of f (x) = 3x4 − 4x3 − 12x2 + 5.

Example

Sketch the graph of f (x) = x2−4x2−1

.

Applications of differentiation