The bad news: this is not a polynomial function, it is a polynomial (#3x+7#) minus a trigonometric function #2sinx#).
The good news is that #f# is a continuous function, because each of its terms are continuous. (Continuous on its domain, which is all of #RR#.)
Maybe bad news, we cannot use the intermediate value theorem to show that the function has EXACTLY one zero. The good news is that we can use it to show that the function has at least one zero.
Proof using the Intermediate Value Theorem
#f# if continuous on the interval #[-pi,0]# (remember it is continuous everywhere.
#f(-pi) = -3pi+7# is negative and #f(0) = 7# is positive.
#0# is between #f(-pi)# and #f(0)#, so, by the Intermediate Value Theorem, there is at least one number #c# in #[-pi,0]# with #f(c)=0#.
Bonus good news
We can use Rolle's Theorem or the Mean Value Theorem to show that the function has at most one zero.
If #f# had two (or more) zeros call them #a# and #b#.
Note that #f# is continuous on #[a,b]# (both the linear and the trigonometric terms are continuous everywhere) and
#f# is differentiable on #(a,b)#. (In fact, #f'(x) = 3-2cosx# is defined for all #RR#, but we only need the interval.)
Then by either Rolle's or the Mean Value there would be a #c# (in#(a,b)#) with #f'(c)=0#
That is: if #f# had two (or more) zeros then #f'(x)=0# would have to have a (real) solution.
BUT #f'(x) = 3-2cosx = 0# has no (real) solutions. (It would require #cosx = 3/2 > 1#.)
Therefore, #f# cannot have two (or more) zeros.
