Exercise 1: Is x^4 Convex Everywhere? — Possible Solution
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STEP 1: COMPUTE THE SECOND DERIVATIVE
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f(x) = x^4
f'(x) = 4x^3 (power rule)
f''(x) = 12x^2 (power rule again)

STEP 2: CHECK THE SIGN AT x=-2, 0, 2
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x=-2: f''(-2) = 12*(-2)^2 = 12*4 = 48. Positive.
x=0:  f''(0)  = 12*0^2 = 0. Zero (neither positive nor negative).
x=2:  f''(2)  = 12*2^2 = 12*4 = 48. Positive.

RESULT
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f''(x) = 12x^2 is never negative anywhere (it's a square,
multiplied by a positive constant, so it can only be zero or
positive) - per this chapter's own test (f''(x) >= 0 everywhere
means convex), x^4 IS convex everywhere.

The one interesting subtlety: at x=0 specifically, f''(0)=0 exactly
(not strictly positive) - this is still consistent with convexity
(the test only requires f''(x) >= 0, not strictly greater than 0),
but it does mean x=0 is a slightly "flatter" point on the curve
than a typical convex function like x^2 would have (whose own
second derivative is a constant 2, never touching zero).

WHY THIS WORKS AS AN ANSWER
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The second derivative is computed explicitly via the power rule
applied twice, checked at all three requested points individually,
and the borderline case (f''(0)=0 exactly) is addressed honestly
rather than glossed over, since the convexity test's own ">=0"
wording specifically allows for this.