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Relation and FunctionsQuestion and Answers: Page 2

Question Number 204478 Answers: 1 Comments: 0

soit f: R^3 →R^3 f(x,y,z)=(x+y,2x−y,x+z) •1 Ecrire la matrice M de cette application dans la base canonique B de R^3 •2 Calculer f(1,2,3)de 2 manieres differentes −en utilisant la definition de f −en utilisant la matrice M •3 determiner bsse de Ker( f) et de Im(f) •4 soient v_1 =(1,1,0) v_2 =(1,2,1) v_3 =(1,3,1) Montrer que la famille E=(v_1 , v_2 , v_3 )est une base de R^3 •5Calculer f(v_1 ) donner ses coordonnes(locus) dans bass E avec f(v_2 )=v_1 +6v_2 −4v_3 f(v_3 )=2v_1 +8v_2 −6v_3 •6 Ecrire la matrice N de f dans base F •7 Retrouver cette matrice a partir de M en utilisant la formule de changement de base

$soit f : R^{3} \to R^{3} f (x, y, z) = (x + y, 2 x - y, x + z)$ $∙ 1 Ecrire la matrice M de cette application$ $dans la base canonique B de R^{3}$ $∙ 2 Calculer f (1, 2, 3) de 2 manieres differentes$ $- en utilisant la definition de f$ $- en utilisant la matrice M$ $∙ 3 determiner bsse de Ker (f) et de I m (f)$ $∙ 4 soient v_{1} = (1, 1, 0) v_{2} = (1, 2, 1) v_{3} = (1, 3, 1)$ $Montrer que la famille E = (v_{1}, v_{2}, v_{3}) est$ $une base de R^{3}$ $∙ 5 Calculer f (v_{1}) donner ses coordonnes (locus)$ $dans bass E$ $avec f (v_{2}) = v_{1} + {6 v}_{2} - {4 v}_{3}$ $f (v_{3}) = {2 v}_{1} + {8 v}_{2} - {6 v}_{3}$ $∙ 6 Ecrire la matrice N de f dans base F$ $∙ 7 Retrouver cette matrice a partir de M$ $en utilisant la formule de changement de base$

Question Number 204426 Answers: 1 Comments: 0

let a , b >0 find all differentiable function f:(0,∞)→(0,∞) such that f′((a/x)) = ((bx)/(f(x))) , ∀ x>0

$let a, b > 0 find all differentiable function$ $f : (0, \infty) \to (0, \infty) such that$ $f^{'} (\frac{a}{x}) = \frac{bx}{f (x)}, \forall x > 0$

Question Number 204394 Answers: 0 Comments: 0

Question Number 204393 Answers: 0 Comments: 0

Question Number 203905 Answers: 0 Comments: 8

let ABC a given triangle. Can we find three positions I,J,K on the side AB,AC,BC Such that IJK is equilateral?

$l e t A B C a g i v e n t r i a n g l e . C a n w e f i n d$ $t h r e e p o s i t i o n s I, J, K o n t h e s i d e A B, A C, B C$ $S u c h t h a t I J K i s e q u i l a t e r a l ?$

Question Number 203291 Answers: 1 Comments: 0

f(x)={1+((√x^2 )/x) if x#0 2 if x=0 study the continuty of f in 0

$You can't use 'macro parameter character #' in math mode$ $2 i f x = 0$ $s t u d y t h e c o n t i n u t y o f f i n 0$

Question Number 202865 Answers: 3 Comments: 1

find the seauence u_n wich verify u_0 =1 and u_n +u_(n+1) =(((−1)^n )/n) for n≥1

$f i n d t h e s e a u e n c e u_{n} w i c h v e r i f y$ $u_{0} = 1 a n d u_{n} + u_{n + 1} = \frac{{(- 1)}^{n}}{n}$ $f o r n ⩾ 1$

Question Number 202795 Answers: 3 Comments: 0

if f(−1)=f(0)=f(2)=0 and f(1)=6 then find f(x)=?

$i f f (- 1) = f (0) = f (2) = 0 a n d f (1) = 6$ $t h e n f i n d f (x) = ?$

Question Number 202536 Answers: 2 Comments: 0

Question Number 201848 Answers: 1 Comments: 0

Question Number 201681 Answers: 1 Comments: 0

f(x+1)−f(x)=3f(x)×f(x+1) D_f =N 2023×f(1402)=1 have equation f(x)=1 solution?

$f (x + 1) - f (x) = 3 f (x) \times f (x + 1)$ $D_{f} = N$ $2023 \times f (1402) = 1$ $h a v e e q u a t i o n f (x) = 1 s o l u t i o n ?$

Question Number 200973 Answers: 0 Comments: 0

Question Number 200722 Answers: 1 Comments: 1

Find all polynomials P(x) with real coefficients such that for all nonzero real numbers x, P(x)+P((1/x))=((P(x+(1/x))+P(x−(1/x)))/2)

$Find all polynomials P (x) with$ $real coefficients such that for$ $all nonzero real numbers x,$ $P (x) + P (\frac{1}{x}) = \frac{P (x + \frac{1}{x}) + P (x - \frac{1}{x})}{2}$

Question Number 199781 Answers: 1 Comments: 0

Question Number 199167 Answers: 0 Comments: 0

Give a function f: R→(0;+∞) continous on R and such that f(x+y) = f(x).f(y) a. Prove f(0) = 1 b. Let h(x) = ln[f(x)]. Prove that: h(x+y) = h(x) + h(y) c. Find all the function f such that problem request

$G i v e a f u n c t i o n$ $f : R \to (0; + \infty) c o n t i n o u s o n R a n d s u c h t h a t$ $f (x + y) = f (x) . f (y)$ $a . P r o v e f (0) = 1$ $b . L e t h (x) = l n [f (x)] . P r o v e t h a t :$ $h (x + y) = h (x) + h (y)$ $c . F i n d a l l t h e f u n c t i o n f s u c h t h a t p r o b l e m r e q u e s t$