 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2022/discrete-mathematics-continuous-mathematics-and-probability-resit/6/b/solution

Substitution gives

f (1) = - 6, f^{'} (1) = f^{''} (1) = f^{(3)} (1) = 0, f^{(4)} (1) = 24.

In particular,

1

x

sufficiently close to

1

, Taylor's theorem with fifth-order remainder gives

f (x) = - 6 + (x - 1)^{4} + \frac{f ^{(5)} ( ξ )}{120} (x - 1)^{5},

where

ξ

lies between

1

and

x

. The fifth derivative is bounded near

1

, so the remainder is

O ((x - 1)^{5})

and is smaller in magnitude than half the positive quartic term for sufficiently small nonzero

x - 1

. Therefore $x = 1$ is a strict local minimum, despite the vanishing second derivative.

An exact algebraic check gives the stronger global conclusion:

f (x) + 6 = \frac{( x - 1 ) ^{4}}{x ^{2}} \geq 0 (x \neq = 0) .

Equality holds only at

x = 1

, so it is the unique global minimum on the actual domain.

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