consider any two closed intervals of R, say [a,b] and [c,d]. show?

Question

that they have the same cardinality by exhibiting an explicit linear function ( i.e. a function of the form f(x) = alpha x + beta) from [a,b] to [c,d] which is bijective.

Answer 1

We want a function f(x) = mx + n so that f(a) = c and f(b) = d. Substituting a and b into the function formula we get
c = f(a) = ma + n
d = f(b) = mb + n.
Solving for m and n, we get
d - c = mb - ma
m = (d - c) / (b - a) and
n = (bc - da) / (b - a)

Substituting these formulas for m and n, we get

f(x) = ((d - c) / (b - a)) x + (bc - da) / (b - a)

Answer 2

It suffices to set f(x) = (d-c)/(b-a) (x-a) + c, or if your teacher insists on expanded form, f(x) = αx+β where α = (d-c)/(b-a) and β = -a(d-c)/(b-a) + c. It is relatively easy to show that this gives a bijection from [a, b] to [c, d].