Diketahui f(x)=3^2x-1.tentukan nilai dari

Diketahui f(x)=3^2x-1.tentukan nilai dari

F(x)=3^(2x-1)

f(x – 1) = 3^(2(x-1) – 1)
= 3^(2x -3)
=3^2x / 3^3

f(x + 1) = 3^(2(x+1) – 1)
= 3^(2x + 1)
=3^2x × 3
= 3× 3^2x

f(x + 2) = 3^(2(x+2) – 1)
= 3^(2x + 3)
=3^2x × 3^3
= 3^3 × 3^2x

jadi
[ f(x-1) + f(x+1) ] / f(x+2)
= [ (3^2x / 3^3) + (3× 3^2x) ] / ( 3^3 × 3^2x)
= 3^2x { [1/27 + 3] / 27 }
= 3^2x {82/27^2}
= 82/729 × 3^2x

$f(x) = {3}^{2x - 1}$
$f(x - 1) = {3}^{2x - 1} \ f(x - 1) = {3}^{2(x - 1) - 1} \ f(x - 1) = {3}^{2x - 2 - 1} \ f(x - 1) = {3}^{2x - 3} \ f(x - 1) = frac{ {3}^{2x} }{ {3}^{3} }$
$f(x + 1) = {3}^{2x - 1} \ f(x + 1) = {3}^{2(x + 1) - 1} \ f(x + 1) = {3}^{2x - 1} \ f(x + 1) = frac{ {3}^{2x} }{3}$
$f(x + 2) = {3}^{2x - 1} \ f(x + 2) = {3}^{2(x + 2) - 1} \ f(x + 2) = {3}^{2x + 3} \ f(x + 2) = {3}^{2x}. {3}^{3}$
$frac{f(x - 1) + f(x + 1)}{f(x + 2)} \ = frac{ frac{ {3}^{2x} }{ {3}^{3} } + {3}^{2x} .3 }{ {3}^{2x} . {3}^{3} } \ = {3}^{2x} frac{ frac{1}{27} + 3 }{27} \ = {3}^{2x} ( frac{ 82}{ {27}^{2} } ) \ = frac{82}{729} . {3}^{2x} \$