f(x)=a0+a1x+a2x^2+a3x^3+a4x^4

f(0)=1, f(1)=3, f(2)=9, f(3)=27, f(4)=81

I think that 3^x and 0<x<5 are intended. a0 to a4 are coefficients.

We need the polynomial to reproduce the series 1, 3, 9, 27, 81.

a0=1 when x=0. The letters A to J represent the equations they label.

A: f(1)=3=1+a1+a2+a3+a4

B: f(2)=9=1+2a1+4a2+8a3+16a4

C: f(3)=27=1+3a1+9a2+27a3+81a4

D: f(4)=81=1+4a1+16a2+64a3+256a4

E: B-2A: 3=-1+2a2+6a3+14a4

F: C-3A: 18=-2+6a2+24a3+78a4

G: F-3E: 9=1+6a3+36a4

H: D-4A: 69=-3+12a2+60a3+252a4

I: H-6E: 51=3+24a3+168a4

J: I-4G: 15=-1+24a4, a4=16/24=2/3

From G: 6a3=8-36(2/3)=-16, a3=-16/6=-8/3

From F: 6a2=20+24(8/3)-78(2/3)=20+64-52=32, a2=32/6=16/3

From A: a1=2-(a2+a3+a4)=2-(16/3-8/3+2/3)=2-10/3=-4/3

f(x)=1-4x/3+16x^2/3-8x^3/3+2x^4/3

This polynomial satisfies the requirements of f(x)=3^x for 0<x<5.