Exercise 1: Numerical Derivative of x^3 at x=2 — Possible Solution
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GIVEN
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f(x) = x^3, x = 2
True derivative: f'(x) = 3x^2, so f'(2) = 3*(2^2) = 3*4 = 12

STEP 1: h=0.1
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f(x+h) = f(2.1) = 2.1^3 = 9.261
f(x) = f(2) = 8
approx = (9.261 - 8) / 0.1 = 1.261 / 0.1 = 12.61

error = |12.61 - 12| = 0.61

STEP 2: h=0.001
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f(x+h) = f(2.001) = 2.001^3 = 8.012006...
f(x) = 8
approx = (8.012006... - 8) / 0.001 = 12.006...

error = |12.006 - 12| = 0.006

RESULT
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h=0.1 gives approx=12.61, error=0.61
h=0.001 gives approx=12.006, error=0.006

The error shrank by roughly a factor of 100 when h shrank by a
factor of 100 (from 0.1 to 0.001) - confirming the same convergence
pattern this chapter's own x^2 example showed, now on a different
function.

WHY THIS WORKS AS AN ANSWER
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Both approximations are computed directly using this chapter's own
numerical differentiation formula, the true derivative is computed
independently via the actual power rule (stated, not yet proven,
per this chapter's own honest framing), and the shrinking error
trend is confirmed numerically rather than just assumed to hold for
a different function than the chapter's own worked example.