How to find the absolute minimum and maximum values

Sharouq Mohammed - 201005184

How to find the absolute minimum and maximum

values

4.1

Sharouq Mohammed - 201005184

The closed interval method:

Continues function on a closed interval [a,b]

1)Find the values of f at the critical numbers of f in (a,b).2)Find the end values of the end points of the interval .

The largest of these values is the absolute maximum the smallest these values is the absolute minimum

Sharouq Mohammed - 201005184

Example:Find the absolute maximum and minimum values 4 ≤ ≤ of the function f(x)=

First we get f’(x)=0 f’(x)= -6

f’(x)=3x(x-2)=0

Critical points : X=0 , x=2

The endpoints: x=4 , x=

Sharouq Mohammed - 201005184

- 4 2 0x

F(x)

The absolute maximum

The absolute minimum

1 -3 17𝟏𝟖

Sharouq Mohammed - 201005184

Sharouq mohammed

Do you have any question?

Hanan Al-Ali200903572

The Mean Value Theorem And The Rolle's Theorem

Hanan Al-Ali (200903572)

If f(x) is1) Continuous on [a , b ]2) Differentiable on (a , b)Then:Exists c in [a , b ] such that:f’(c)= [f(b) – f(a)] /( b – a) And this is (the slope of the line between (a,f(a)) and (b, f(b))

The Mean Value theorem

Hanan Al-Ali (200903572)

f(x) = x2 – 2x – 3 ; [ 0,1]f(x) is continuous because it’s a polynomial and differentiable on[ 0,1]Because a polynomial is differentiable on RSo exists c such that:f ’(c) =2c – 2 = [f(1) – f(0)] /( 1-0) = (1-2-3+3)/1=-1/12c – 2 = -1 2c = -1 + 22c = 1c= ½ € [-2 , 3]

So our solution is ½

Example:

Hanan Al-Ali (200903572)

It a special case of the mean value theorem when: same conditions as the MVT plus f(a)= f(b)

The conclusion: Exists c in (a,b)f’(c)= 0 The slope of the line between (a,f(a)), (b,f(b)) is 0.

The Rolle’s Theorem

Hanan Al-Ali (200903572)

f(x) = x2 – 2x – 3 ; [-1, 3] f(x) is continuous because it’s a polynomialf’(x)= 2x – 2 so it’s differentiable f(-1)= 0f(3)= 0f’(c)= 0 2c – 2 =0 c= 1 € [-1,3] so this is our solution

Example:

Hanan Al-Ali (200903572)

4.3How the derivative effect on the graph

Sara Mohamed Tawfik201001917

Sara Moursi - 201001917

Increasing and Decreasing

c

• The tangent line between a and b , c and d have a +ve Slope f’(x)>0• The tangent line between b d b and c have a –ve slopeF’(x)<0 Note :• F increases when f’(x) is +ve• F decreases when f’(x) is -ve

Sara Moursi - 201001917

Local maxima and minima

• Where can we find the min and max? Local min/max must occur at a place in the domain where

f switches from increasing to decreasing or vice versa, we can check where the min/max from the critical number.

If f’(x) negative to the left of c, f’(x) positive to the right of c (local-min at c).

If f’(x) positive to the left of c, f’(x) negative to the right of c (local-max at c).

If the sign is the same on sides of c, then its neither min nor max.

Sara Moursi - 201001917

Concavity

• How can we know the concavity? We can know the concavity through the

second derivative f’’(x)• If f’’(x) is positive then the function is concave

UP.• If f’’(x) is negative then the function is concave

DOWN.

Sara Moursi - 201001917

Inflection points

• A point (x, y) on the graph of the function is called inflection point if f switches concavity on x

• If f(x) switches concavity at x=0 and f(x) is undefined at 0 then there is no inflection points.

Example: f(x) = 1/x.

Sara Moursi - 201001917

Example

Show that the curve y= x^4 – 4x^3 have concavity., points of inflection and local maxima and minima

Solution :1) Find the f’(x) 2) Take common factor3) Take f’(x) = 0 (to find critical points)4) Get f’’(x)5) Substitute the critical points in f’’(x) , f’(x) and f(x)

to get the local min or max (local min at f(3)

Sara Moursi - 201001917

Sara Moursi - 201001917

Concave upwards and downwards

Sara Moursi - 201001917

Point of inflection and Graph

• (0,0) is the inflection point since the curve changes from concave upwards to concave downwards.• (2,-16) is an inflection point since the curve changes from concave downward to upward

Section 4.4limit at infinity: horizontal and vertical asymptote

Amna Ahmad200914882

Amna Ahmad- 200914882

Amna Ahmad- 200914882

• The horizontal asymptote is when the x goes to infinity and vertical asymptote is when the function to a value that the result of the limit is infinity.

Horizontal asymptote

Vertical asymptote

Amna Ahmad- 200914882

Amna Ahmad- 200914882

Example 2: find the horizontal and vertical asymptotes of the graph of the function:

• F(x)=

• The root of x square is equal the absolute value of x.

• The absolute value of x is equal +xe if x is positive and –x if x is negative.

Amna Ahmad- 200914882

F(x)=

=

=

Amna Ahmad- 200914882

So y= is horizontal asymptote of the graph f

=

=

Amna Ahmad- 200914882

And when x goes to negative infinity :

=

=

Amna Ahmad- 200914882

Thus y = is also horizontal asymptote.

A vertical asymptote is occurring when the denominator, 3x-5 is 0. If x is close to 5/3 and x> 5/3 , then the dominator is close to 0 and 3x-5 is positive . The numerator is always positive, so f(x) is positive.

=

If x is close to 5/3 but x<5/3 , then 3x-5<0 and so f(x) is large negative. Thus

=

-

Amna Ahmad- 200914882

The vertical asymptote is x= 5/3.