Oxford Entrance Exam Problem 3

Calculus Level 3

The cubic y = k x 3 − ( k + 1 ) x 2 + ( 2 − k ) x − k y=kx^3-(k + 1)x^2+(2-k)x-k has a turning point, that is a minimum, when x = 1 x = 1 precisely for

k < 3 k<3 0 < k < 1 0<k<1 k > 1 2 k>\frac{1}{2} k > 0 k>0 All values of k k

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1 solution

When a function y = f ( x ) y = f(x) has a minimum at x = a x=a , then d 2 y d x 2 > 0 @ x = a \frac{d^2y}{dx^2} > 0 \text{ @ } x = a

We apply the same logic to this problem.

f ′ ′ ( x ) = 6 k x − 2 ( k + 1 ) x f''(x) = 6kx - 2(k+1)x

f ′ ′ ( 1 ) = 6 k − 2 k − 2 > 0 ⇒ k > 1 2 f''(1) = 6k - 2k - 2 > 0 \Rightarrow k > \boxed{\frac{1}{2}}

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