Page 291 - MATLAB an introduction with applications

276 ——— MATLAB: An Introduction with Applications

2
Let F(X) = a + a X + a X . From the given data, we have
2
0
1
a + a + a 2 = –7
0
1

a + 2a + 4a = 5
2
1
0

a + 3a + 9a = 14
1
0
2
Solving above equations give
a = –22
0
a = 16.5
1
a = –1.5
2
>> x=[1 2 3];y=[–7 5 14];a2a1a0=polyfit(x,y,2)
a2a1a0 =
–1.50000000000000 16.50000000000002 –22.00000000000003
and the function is
F(X) = –22 + 16.5X – 1.5X 2

dF ()
X
−
For finding minimum of the function, we find the first derivative of the function = 16.5 3X = ;
0
dX
for a minimum or maximum
thus, X = 5.5
2
()
dF X
<
To check for minima or maxima, we find second derivative of the function =− 30 ; thus it’s a
dX 2
maximum.
Thus, for the given function there is no absolute minimum. The function attains its minimum values at ±∞ .
MATLAB Solution:
% Problem 3.2
clc
clear
disp(‘Fit a Polynomial by Quadratic Aproximation’)
x1=1;
x2=2;
x3=3;
f1=–7;
f2=5;
f3=14;
fx1=[1 x1 (x1)];
fx2=[1 x2 2*x2];
fx3=[1 x3 3*x3];
a=[fx1;fx2;fx3]
b=[f1 ;f2; f3]
poly_values=inv(a)*b