how to find the absolute minimum and maximum values

Sharouq Mohammed - 201005184

How to find the absolute minimum and maximum

values

4.1


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


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=


- 4 2 0x

F(x)

The absolute maximum

The absolute minimum

1 -3 17𝟏𝟖


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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


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:


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


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:


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


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.


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.


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.


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)



Concave upwards and downwards


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

• 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



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.


F(x)=

=

=


So y= is horizontal asymptote of the graph f

=

=


And when x goes to negative infinity :

=

=


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

=

-


The vertical asymptote is x= 5/3.

how to find the absolute minimum and maximum values

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