diketahui rumus suatu fungsi f(x) =ax +b f(3) = 5 dan f(-1) = 3. Nilai f(4) adalah

f(x) = ax+b
f(3)=5 ⇔ 3a + b = 5...(i)
f(-1)=3⇔ -a+b = 3...(ii)

3a+b = 5
-a +b = 3
------------- -
4a = 2
a = 2/4
a = 1/2
substitusikan nilai a ke pers (ii):
-a + b = 3
-(1/2)+b = 3
b = 3+1/2
b = 7/2
sehingga:
f(4) = 4a+b
f(4) = 4(1/2) + (7/2)
f(4) = 4/2 + 7/2
f(4) = 11/2

semoga membantu