diketahui rumus fungsi adalah f(2x + 1)=4x + 5 nilai f(2) adalah...

f (2x + 1) = 4x + 5

f (2x + 1) = f (2)
2x + 1 = 2
2x = 2 - 1
2x = 1
x = 1/2

Nilai f (2)
= 4x + 5
= 4 (1/2) + 5
= 2 + 5
= 7

f(2x+1) = 4x+5
f(2) = . . .
2x+1 = 2
2x = 2-1
x = ½
f(2) = f(2.½+1) = 4.½+5 = f(2) = 7