Practice problem 3
Problem 3(a)
For testing: we compute
:
clearvars
q1 = g(0)
q1 =
0
q2 = g(1.5)
q2 =
0.856188393624901
Problem 3(b)
The function
is continuous, so we can use the intermediate value theorem.
From (a) we know that there must be a value
such that
.
We define
(see below).
as = fzero(@G,[0,1.5])
% must use @G
as =
0.551039427609027
Definition of functions g, G
All function definitions must be at the end of the mlx file.
function
q = g(a)
f = @(t) exp(-t.^2);
q = integral(f,0,a);
end
function
q = G(a)
q = g(a)-0.5;
end