Question

Fungsi f dinyatakan dengan rumus f[x]=px+ q. diketahui f[3]= 11 dan f[1]= 7. nilai p dan q adalah

Answer 1

F(x) = px + q, mka f(3) = px+q,
 px +q = 11
p(3)+q = 11
3p+q = 11 ⇒ P1

f(1) = px+q
 px +q = 17
p(1) + q = 17
p+q = 17
p+q = 17
                 ⇒ P2

ELIMINASI

3p+q = 11
p+q = 17 -
2p = - 6
p = -3

substitusikan p = -3 ke p1
f(3) = px+q = 11
-3.3 +q = 11
-9 +q = 11
-9+9+q = 11+9
q =20


maaf kalo salah ya..

Answer 2

F(3) = a(3) + b = 11 -> = 3a + b = 11
f(1) = a(1) + b = 7 -> =  a + b = 7      -
                               -----------------------
*lalu dieliminasikan b nya
= 2a  = 4
a = 2
f(3) = 3a + b = 11
= 3(2) + b = 11
= 6 + b = 11
= b = 11 - 6
= b = 5