Exercise 2: Proving f(n) = 2n + 7 Is Theta(n) — Possible Solution
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GIVEN
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f(n) = 2n + 7

STEP 1: THE O(n) BOUND, c=3
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Testing f(n) <= 3n at successive values of n:

  n=6: f(6) = 2(6)+7 = 19.  3n = 18.  19 <= 18? NO.
  n=7: f(7) = 2(7)+7 = 21.  3n = 21.  21 <= 21? YES (exactly equal).

So the smallest n0 for which f(n) <= 3n holds for all n >= n0 is:

  n0 = 7

STEP 2: THE OMEGA(n) BOUND
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Using c=2: f(n) = 2n + 7 >= 2n holds for EVERY n >= 1, since the extra
+7 term is always positive and can only make the left side larger,
never smaller, than 2n. So:

  Omega(n) holds with c=2, n0=1

STEP 3: CONCLUDING THETA(n)
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Since both f(n) = O(n) (with c=3, n0=7) and f(n) = Omega(n) (with c=2,
n0=1) are proven with valid constants, per this chapter's own
definition:

  f(n) = Theta(n)

WHY THIS WORKS AS AN ANSWER
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Both bounds are proven separately with explicit, verified constants -
the O(n) bound's n0 is found by checking successive integers until
the inequality first holds, and the Omega(n) bound is shown to hold
trivially for all n due to the always-positive +7 term - and the
Theta conclusion is drawn only after both individual proofs are
established, exactly per this chapter's own definition of Theta as
requiring both O and Omega simultaneously.