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Unformatted text preview: FUNCTIONS I- SOME DEFINITIONS A function defined from to is a mapping which to every element of corresponds at most an element of . We denote : ⟶ ⟼ () () is the image of the object If to every element of is associated exactly one element of , it is called an application. The Domain of definition of a function : ⟶ denoted or are the set of values of having images in . = { ∈ , () } 1 Example : The function defined by () = exist iff ≠ 0. So =] − ∞, [ ⋃], + ∞[ The graph of is the set () = {(, ()), ∈ } Let and be two functions defined from to . ∀ ∈ , ( + )() = () + () ∀ ∈ , ()() = () × () ∀ ∈ , ()() = () If ≠ 0, ∀ ∈ , ()() = () + = . = ⋂ / = ⋂ ⋂{ ∈ , () ≠ 0} ∘ = ∩ −1 ( ) () When = ℝ, we talk of a numeric function. When = = ℝ, we talk of a numeric function with one real variable. Parity and periodicity of a function: A function is said to be : Even if is symmetric about the origin and (−) = (). i.e. ∀ ∈ , (−) ∈ and (−) = () The curve of an even function is symmetric about the ordinate axis. Odd if is symmetric about the origin and (−) = −(). i.e. ∀ ∈ , (−) ∈ and (−) = −() The curve of an odd function is symmetric about the origin. Periodic if there exist ∈ ℝ∗+ such that ∀ ∈ , ( + ) ∈ and ( + ) = () The smallest value of ∈ ℝ∗+ is called the period. A function : ⟶ is said to be : Lower bounded if ∃ ∈ ℝ, ∀ ∈ , () ≥ Upper bounded if ∃ ∈ ℝ, ∀ ∈ , () ≤ Bounded if it is both upper bounded and lower bounded. i.e. ∃, ∈ ℝ, ∀ ∈ , ≤ () ≤ i. e. ∃ ∈ ℝ∗+ , ∀ , |()| ≤ Injective if two distinct elements of always have distinct images. That is ∀, ∈ , ≠ ⟹ () ≠ () (ie. () = () ⟹ = ) Surjective if every element of has at least one object in . ∀ ∈ , ∃ ∈ such that = () Bijective if it is both injective and surjective. That is: ∀ ∈ , ∃! ∈ such that = () The inverse (bijective inverse) of is the function denoted − is such that: ∀ ∈ , −1 () ∈ i.e. = −1 () ⟺ = () NB: The notation −1 has a meaning only when is a bijection. Furthermore, is bijective iff −1 is bijective. That is ( −1 )−1 = . Increasing (resp. decreasing) on iff ∀, ∈ , ≤ ⟹ () ≤ () (resp. () ≥ ()). II- Strictly increasing (resp. strictly decreasing) iff ∀, ∈ , < ⟹ () < () (resp. () > ()) STUDYING THE LIMITS OF A FUNCTION Given a function : ℝ ⟶ ℝ ↦ () Studying the limit of when tends to 0 is determining the value (finite or infinite) towards which tends. Some Standard Limits. lim (ln ) = 0 (, ) ∈ ℝ∗+ × ℝ ⟶0+ (ln ) =0 ⟶+∞ lim = +∞ ⟶+∞ − lim = ⟶+∞ sin lim = 1 ⟶+∞ lim 0 (, ) ∈ ℝ × ℝ∗+ lim ln = 0 ⟶0+ ln(1+) lim ⟶0 =1 Exercice 1 Given the following functions, state their domain of definitions, calculate the limit at their boundaries, comment on their monotonicity and say whether they are lower bounded, upper bounded, even, odd or periodic. a) () = 7 5 +4 3 −1 2 ++6 2−1 1 b) () = √ +1 c) ℎ() = sin 1 d) () = ( ) (limit at 0 and at +∞) Operations on limits Given two functions and defined on ℝ. lim ( + ) = lim () + lim () ⟶0 ⟶0 ⟶0 lim () = lim () × lim () ⟶0 ⟶0 ⟶0 lim () = lim () ⟶0 lim ( ) ⟶0 ⟶0 lim () = ⟶0 lim () ⟶0 where lim () ≠ 0 ∀ ⟶0 ̅. Let , ∈ ℝ and 0 ∈ ℝ () +∞ −∞ +∞ −∞ +∞ () +∞ −∞ −∞ ( + ) + +∞ −∞ +∞ −∞ ? ( ∙ ) +∞ > 0 {−∞ < 0 ? else +∞ < 0 {−∞ > 0 ? else +∞ +∞ −∞ +∞ −∞ ⟶ ⟶ ⟶ ⟶ ( )() ⟶ ≠ 0 +∞ = 0, > 0 −∞ = 0, < 0 ? = = 0 { 0 = ±∞ ? ? ? For simplicity, we denoted ≪ ? ≫ in the table above to mean ≪ undefine ≫. But practically, when confronted to such difficulties, we further simply the expressions. NB: Some limits exist (finite, infinite or undefined) and others don’t. Given two functions , and ℎ defined on ℝ. If () ≤ () (resp. () ≥ ()) then lim () ≤ lim () (resp. lim () ≥ lim ()) ⟶0 ⟶0 ⟶0 ⟶0 If () ≥ () and lim () = +∞ then lim () = +∞. If () ≤ () and lim () = −∞ then lim () = −∞. ⟶0 ⟶0 ⟶0 ⟶0 If () ≤ () ≤ ℎ() then lim () ≤ lim () ≤ lim ℎ(). (squeeze theorem) ⟶0 ⟶0 ⟶0 Furthermore, if and ℎ have the same limit say at 0 , then has a limit at 0 . Limits of composite functions Given two functions and such that lim () = and lim () = ′ ⟶0 ⟶ Then : ( ∘ ) = ′ ⟶0 Image of a convergent sequence ̅ at the point Let be a function defined on an interval and ∈ . has a limit ∈ ℝ iff for every sequence ( ) converging towards , the sequence (( )) converges towards . ( lim () = ) ⇔ (∀ sequence ( ) lim = , lim ( ) = ) ⟶ ⟶+∞ ⟶+∞ Consequently, in order to show that a function has no limit when ⟶ , we may just find a sequence ( ) that converges towards such that the sequence (( )) is instead divergent. Exercise 2 Calculate the following limits: 1 1) sin at 0. 2) 3) 4) 5) 6) 7) 8) 1 lim sin ⟶+∞ lim (√5 − 2 − √ + 1) ⟶+∞ lim (√5 − − √1 − ) ⟶−∞ lim ( − √1 + 2 ) ⟶+∞ lim sin ⟶+∞ 2 +1 sin() lim (tan ) ⟶0 lim 3 (2 + ⟶+∞ cos ) III- CONTINUITY OF A FUNCTION A function : ℝ ⟶ ℝ is said to be continuous at a point 0 if () = () = ( ) → → < > That is: () = ( ) → If () = ( ) (respectively () = ( )) we say is continuous on the → → < > left (resp. on the right) of . Given an interval and 0 ∈ ℝ such that − { 0 } ≠ ∅ and : − { 0 } ⟶ ℝ a continuous function on − { 0 } such that lim () = ∈ ℝ. Then the extension ⟶0 ( ≪prolongement par continuité ≫) of the function at 0 is the function defined by () ∀ − { } ̃() = { = Given ⊂ ℝ. is continuous on if is continuous at all points of . Given and two continuous functions on an interval then + , − , N.B , , are all continuous on . Intermediate Value Theorem: Let be a continuous function on [, ], ≠ and let 0 ∈ [(), ()], then: ∃ ∈ [, ] such that () = 0 . In particular, for 0 = 0, it means that: if is continuous on and there exist , ∈ such that () and () are of different signs (i.e. () × () < 0 ), then the equation () = 0 admits atleast one solution on . Extreme Value Theorem: On a closed interval, a continuous function attains its maximum and its minimum. Let be a continuous function on an interval . ( ) ⇔ ( ) Note If the function : ⟶ ℝ is continuous and strictly monotonous on , then is a bijection between and = (). It thus has an inverse −1. This inverse is also continuous and strictly monotone (same direction as ) between = () and . Recall : Inverse Trigonometric and hyperbolic functions ( ∈ [− , ] and sin = ) ⟺ ( ∈ [−1,1 ] and arcsin = ) 2 2 ( ∈ [0, ] and cos = ) ⟺ ( ∈ [−1,1 ] and arccos = ) ( ∈ [− , ] and tan = ) ⟺ ( ∈ ]−∞, +∞[ and arctan = ) 2 2 ( ∈ ]−∞, ∞[ and Argsh = ) ⟺ ( ∈ ]−∞, +∞[ and sh = ) ( ∈ ]1, ∞[ and Argch = ) ⟺ ( ∈ ]0, +∞[ and ch = ) ( ∈ ]−1,1[ and Argth = ) ⟺ ( ∈ ]−∞, +∞[ and th = ) ∀ ∈ [−1,1], arcsin + arccos = 2 sin(arccos ) = cos(arcsin ) = √1 − 2 > 0 1 arctan + arctan = { 2 − < 0 2 Remarks All polynomial functions are continuous on ℝ. Rational functions, exponential functions, logarithmic functions trigonometric functions and hyperbolic functions are all continuous on their domain of definition. Exercice 3 Find the domain of definition and justify the continuity of the following functions in their domain. (2+1) 1) () = √ 2 −+1 (2+1) 2) () = √| 2 −+1| IV- DERIVABILITY OF A FUNCTION A function : ℝ ⟶ ℝ is derivable (or differentiable) at a point if ⟶ → > Note is finite. This limit is called the derivative of at point and is denoted (). After an appropriate change of variable to the limit expression above, we return to the 1st principle formula we all know. is said to be derivable to the right of (respectively to the left) if : ()−() − ′ ()−() − (resp. → < ()−() − ) exists. is derivable at if it is derivable at the right and left of and if both derivatives at are equal. If is derivable at point , then it is continuous at . (the reverse is false; counter example () = || at 0). If is derivable at point , the curve of admits a tangent at point (, ()) of gradient ′ () and direction vector ⃗(1, ′ ()). The equation of the tangent is given by = ′ ()( − ) + (). Given two derivable functions and then, ( + )′ = ′ + ′ ( × )′ = ′ × + × ′ ()′ = ( ∘ ) () = (()) × ′ () ( −1 )′ () = ′ () ′ ×−×′ 2 ′ ′ 1 1 where = () (i.e. = /) If and are times derivable, ( × )() = ∑= (−) Taylor Young’s series () = (0) + ′ (0) + 2 ′′ (0) + ⋯ + () (0) + ( ) 2! ! Note Given a function defined on an interval . If ′ () > 0 ∀ ∈ then the function is increasing on . If ′ () < 0 ∀ ∈ then the function is decreasing on . If ′ () = 0 ∀ ∈ then the function is constant on . Mean Value Theorem (“theoreme des accroissements finis”) If a function is continuous on an interval [, ] and derivable in ], [, then ∃ ∈ ], [, such that ′ () = ()−() . − Exercise 4: Show that 1. |sin | ≤ ∀ 1 1 +1 ∀ 1 ln(ln 2) + 2ln 2 2. (1 + ) < < (1 + ) ∈ ℕ∗ 3. ≥ ln(ln ) − where = ∑=2 kln . Hence or otherwise deduce the 1 nature of the sequence ( ). Remark: we could determine the nature of ( ) above while using the improper simple integral test for series (view sequences and series) Concave and convex Functions Given a function defined on an interval . + ) 2 is concave if ( ≤ ()+() 2 ∀, . In this case, the curve of is below all its tangents. + ) 2 ()+() 2 is convex if ( its tangents. If is convex on , then is continuous on . is derivable and is convex on iff ′ is increasing on . ≥ ∀, . In this case, the curve of is above all Note Consider a function which is two times derivable on ℝ. V- The point (, ()) such that ′ () = 0 is an extremum. It can either be a maximum or a minimum turning point. If ′′ () = 0, the point (, ()) is a point of inflexion. If ′′ () > 0 (resp. ′′ () < 0 ∀ ∈ then the is locally convex (resp. concave) on . A convex function on an interval is continuous on . STUDYING THE ASYMPTOTES TO A CURVE Given a function : ℝ ⟶ ℝ If has a finite limit at ±∞ then the line = is the horizontal asymptote to the curve of . If has an infinite limit to the left or right of a point , then the line of equation = is the vertical asymptote to the curve of . If [() − ( + )] = then the line = + is an oblique asymptote to ⟶±∞ the curve of . Note The line = + is an oblique asymptote to the curve of iff () lim ⟶+∞ = and lim (() − ) = . ⟶+∞ Direction Asymptote () . ⟶+∞ If () = ±∞ we study lim ⟶±∞ () ⟶±∞ () ⟶+∞ a) If = ±∞ then the − is a vertical asymptote to the curve of . b) If = , ∈ ℝ, then : If = , then the − is a horizontal asymptote to the curve of . If ≠ , we study (() − ). ⟶±∞ if () − has a finite limit say , then the line = + is an asymptote to the curve of . if () − has an infinite limit, the curve of is said to admit a parabolic branch towards the line = . if it has no limit, then the curve is said to have an asymptotic direction = . Exercise 5 For each of the following functions: a) state their domain of definition b) Study their limits at their respective boundaries c) Study the continuity, derivability, monotonicity and state (if any) the asymptotes to the curve. 1) 1 () = 4) sin 4 = 2√ 2 − 1 − VI- 2) 2 = 2 √+1 3) 3 = (−1)2 2 +1 5) Argcoth STUDYING A FUNCTION The following are the steps to follow when asked to completely study a function: 1. 2. 3. 4. 5. State the domain of definition of the function; Precise the continuity of the function on its domain of definition; Calculate the limits at the boundaries of the domain (Asymptotes); Study the parity (to simplify the domain of study) Study the derivability, find the derivative (while stating the domain of derivability) and determine the monotonicity of the function on the different intervals of its domain; 6. Find the turning points (if any); 7. Draw the table of variations of the function; 8. Sketch the curve. Example Study the function () = 3 2 −−2 Solution Definition of the domain The domain of is the set of values for which () exists. () exists if 2 − − 2 ≠ 0 ⟹ ( − 2)( + 1) ≠ 0 =] − ∞; −[ ∪ ] − ; [ ∪ ]; +∞[ is continuous on its domain of definition as a rational function. Calculating limits 3 →−∞ 2 () = →−∞ = −∞ also lim () = +∞ By Euclidean division 3 2 −−2 x→+∞ = ( + 1) + 3+2 2 −−2 And we thus have lim (() − ( + 1)) = 0 ⟶±∞ Thus the line = + is an oblique asymptote to the curve at ±∞. lim f(x) = −∞ lim f(x) = +∞ lim− f(x) = −∞ lim+ f(x) = +∞ x→−1 x→−1 < > x→2 x→2 Then the line = − and = are vertical asymptotes Derivability is derivable on its domain of definition as a rational function. ′ () = 2 ( 2 − 2 − 6) ( 2 − 2 − 2)2 ′ () = 0 ⟺ = 0 = 1 + √7 or = 1 − √7 We determine the sign of ′ () from the polynomial ( 2 − 2 − 6) 2 Since = ℝ./{−1,2}, (2 −2−2)2 ≥ 0 We have;∀ ∈] − ∞; 1 − √7[ ∪ ]1 + √7; +∞[; ′ () > Then f is increasing in ] − ∞; − √[ ∪ ] + √; +∞[ ∀ ∈]1 − √7; −1[ ∪ ] − 1; 2[ ∪ ]2; 1 + √7[; ′ () < Then is decreasing on the interval ] − √; −[ ∪ ]; + √[ Table of Variations 3 (1 − √7) (0) = 0 = (1 − √7) = = −1.89 2 (1 − √7) − (1 − √7) − 2 = 1 − 3√7 + 3(7) − 7 3 3⁄ 2 1 − 2√7 + 7— 1 + √7 − 2 = 22 − 3√7 − (√7) 5 − √7 ≈ 6.34 M and m are maximum and minimum relative to f O is the point of inflexion More Exercises Study the following functions: 3 1. () = 4 √| 2 − 4 − 12| 2 sin 2 2. () = 1+cos 3. Given the function : ]− 2 ; 2 [ ⟶ ℝ 4 2 ⟼ ln(tan ( + )) a. Show that the function is well defined on = ]− 2 ; 2 [. Study . Then show that: b. ℎ () 2 = tan 2 c. ℎ() = ...
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