Question

Va rog,pentru azi.
f:R-> R f(x)=5x-20-3(x+1)-4​

Answer 1

a) Gf intersectat cu Ox = {A}

f(x) = 0 => 5x-20-3(x+1)-4 = 0

5x-24-3x-3 = 0

2x=27

x = 27/2

Deci A(27/2, 0)

Gf intersectat cu Oy = {B}

x = 0 => f(0) = 5*0-20-3(0+1)-4 = -24-3 = -27

Deci B(0, -27)

b) Banuiesc ca aici trebuie prescurtata functia f

f(x) = 5x-20-3(x+1)-4 = 5x-24-3x-3 = 2x-27

c) f(1) = 2*1-27 = 2-27 = -25

f(2) = 2*2-27 = 4-27 = -23

f(3) = 2*3-27 = 6-27 = -21

f(4) = 2*4-27 = 8-27 = -19

f(0) = 2*0-27 = -27

-----------------------------------

-25+23-21+19-27 = -31

Answer 2

funcția f:R→R

$f(x) = 5x-20-3(x+1)-4 = 5x - 20 - 3x - 3 - 4$

$\implies f(x) = 2x - 27$

a)

intersecția cu axa Ox:

$f(x) = 0 \iff 2x - 27 = 0$

$2x = 27 \iff x = \dfrac{27}{2}$

$\implies \Big(\dfrac{27}{2} ; 0\Big)$

intersecția cu axa Oy:

$x = 0 \iff y = 2 \cdot 0 - 27 = -27$

$\implies \Big(0;-27\Big)$

b)

forma ireductibilă:

$f(x) = 2x - 27$

c)

$f(1) = 2 \times 1 - 27 = 2 - 27 = - 25$

$f(2) = 2 \times 2 - 27 = 4 - 27 = - 23$

$f(3) = 2 \times 3 - 27 = 6 - 27 = - 21$

$f(4) = 2 \times 4 - 27 = 8 - 27 = - 19$

$f(0) = 2 \times 0 - 27 = 0 - 27 = - 27$

⇒

$f(1)-f(2)+f(3)-f(4)+f(0) =$

$= - 25 - ( - 23) + ( - 21) - ( - 19) + ( - 27)$

$= - 25 + 23 - 21 + 19 - 27$

$= 42 - 73 = \bf - 31$