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4.1. How to find the absolute minimum and maximum values. 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 .
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4.1 How to find the absolute minimum and maximum values 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 functionf(x)= First we get f’(x)=0 f’(x)= -6 f’(x)=3x(x-2)=0 X=0 , x=2 Critical points : The endpoints: x=4,x= Sharouq Mohammed - 201005184
The absolute maximum 1 17 -3 The absolute minimum Sharouq Mohammed - 201005184
Do you have any question? Sharouqmohammed Sharouq Mohammed - 201005184
The Mean Value Theorem And The Rolle's Theorem Hanan Al-Ali 200903572 Hanan Al-Ali (200903572)
The Mean Value theorem If f(x) is • Continuous on [a , b ] • 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)) Hanan Al-Ali (200903572)
Example: 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 R So exists c such that: f ’(c) =2c – 2 = [f(1) – f(0)] /( 1-0) = (1-2-3+3)/1=-1/1 2c – 2 = -1 2c = -1 + 2 2c = 1 c= ½ € [-2 , 3] So our solution is ½ Hanan Al-Ali (200903572)
The Rolle’s Theorem 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. Hanan Al-Ali (200903572)
Example: f(x) = x2 – 2x – 3 ; [-1, 3] f(x) is continuous because it’s a polynomial f’(x)= 2x – 2 so it’s differentiable f(-1)= 0 f(3)= 0 f’(c)= 0 2c – 2 =0 c= 1 € [-1,3] so this is our solution Hanan Al-Ali (200903572)
4.3How the derivative effect on the graph Sara Mohamed Tawfik 201001917 Sara Moursi - 201001917
Increasing and Decreasing • 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 slope F’(x)<0 • Note : • F increases when f’(x) is +ve • F decreases when f’(x) is -ve c 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 : • Find the f’(x) • Take common factor • Take f’(x) = 0 (to find critical points) • Get f’’(x) • 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
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 Sara Moursi - 201001917
Section 4.4limit at infinity: horizontal and vertical asymptote Amna Ahmad 200914882 Amna Ahmad- 200914882
Vertical asymptote • 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 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. Amna Ahmad- 200914882