The Intermediate Value Theorem (IVT) is a theorem about functions that are continuous on closed interval.
(One version of) IVT says that
If #f# is continuous on #[a,b]# with #f(a) != f(b)# and #M# is between #f(a)# and #f(b)# (exclusive),
then there is a #c# in #(a,b)# with #f(c)=M#
In order to use the IVT, we must have a function that is continuous on a closed interval as the topic of discussion.
One use for the IVT is to show that a function takes on a particular value.
Any solution to #3 = x^2 + sin(pix)# is also a solution to # x^2 + sin(pix)-3=0#, and vice versa.
To use IVT to show that the equation #3 = x^2 + sin(pix)# has a solution, we can
(1) use IVT to show that the function: #f(x) = x^2+sin(pix)# takes on the value #3#
OR
(b) use IVT to show that the function: #f(x) = x^2+sin(pix)-3# takes on the value #0#
I have a slight preference for method (b). So we'll show that #f(x) = x^2+sin(pix)-3# has a (real number) zero.
We now need an interval #[a,b}# on which #f# is continuous. And, since out "target" number #M# is #0#, we want one of #f(a)# and #f(b)# to be positive and the other negative.
For integer #x#, #sin(pix) = 0#, so #f(x) = x^2-3#
#f(1) = -2# is negative and #f(2)= 1# is positive.
Now we're ready to write the proof.
Proof
#x# is a solution to #3 = x^2 + sin(pix)# if and only if #x# is a zero of #f(x) = x^2+sin(pix)-3#.
#f# is continuous on #[1,2]# because it is the sum of continuous functions.
The function, #x^2-3# is a polynomial, so it's continuous everywhere. And #sin(pix)# is the composition of the sine function with #pix# (a linear function), so it is also continuous everywhere.
#f(1) = -1# and #f(2) = 1#, so #0# is (strictly) between #f(1)# and #f(2)#.
Therefore, by the Intermediate Value Theorem, there is a #c# in #(1,21)#, such that #f(c) = 0#.
This #c# is a solution for the equation #3 = x^2 + sin(pix)#.
Important Note
Students sometimes think that the Intermediate Value Theorem tells us how to solve an equation. It does not. It sometimes tells us that a solution must exist, but it does not tell us how to find a solution.