Exercise 2: Finding n0 for f(n) = 2n^2 + 3n + 10 — Possible Solution
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GIVEN
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f(n) = 2n^2 + 3n + 10, c = 3, testing f(n) <= 3n^2

STEP 1: CHECKING SUCCESSIVE VALUES OF n
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n=3: f(3) = 2(9) + 9 + 10 = 18+9+10 = 37.   3n^2 = 3(9) = 27.
     37 <= 27? NO.

n=4: f(4) = 2(16) + 12 + 10 = 32+12+10 = 54.  3n^2 = 3(16) = 48.
     54 <= 48? NO.

n=5: f(5) = 2(25) + 15 + 10 = 50+15+10 = 75.  3n^2 = 3(25) = 75.
     75 <= 75? YES (exactly equal).

STEP 2: CONFIRMING n0 = 5 IS THE SMALLEST VALID VALUE
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The inequality fails at n=3 and n=4, and first holds (as an exact
equality) at n=5. Per this chapter's own formal definition, the
smallest n0 for which f(n) <= 3n^2 holds for ALL n >= n0 is:

  n0 = 5

(Values of n beyond 5 continue to satisfy the inequality too, since
the n^2 term's growing dominance only strengthens as n increases
further past this boundary.)

WHY THIS WORKS AS AN ANSWER
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The inequality is checked at consecutive integer values starting
below the eventual boundary, showing exactly where it transitions from
false to true, and the smallest satisfying n0 is identified precisely
at the point of first success rather than an arbitrary "large enough"
value.