Finding a Number

Calculus Level 3

f ( x ) = x + x x 2 + a f(x)=x+\frac{x}{x^2+a}

For f ( x ) f(x) as defined above, find the largest real value a a for which there exists some real number x 0 x_0 such that f ′ ( x 0 ) = 0 f'(x_0)=0 .

1 8 \frac{1}{8} 1 1 − 3 3 4 \frac{-3\sqrt{3}}{4} 3 2 \frac{\sqrt{3}}{2}

Rafael Resener by Rafael Resener , 23, Brazil

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2 solutions

Differenting the given function with respect to x x and equating with 0 0 , we get x 4 + ( 2 a − 1 ) x 2 + a 2 + a = 0 x^4+(2a-1)x^2+a^2+a=0 . Since there exists a real solution to this equation, the discriminant must be non-negative definite. That is, ( 2 a − 1 ) 2 ≥ 4 ( a 2 + a ) (2a-1)^2\geq {4(a^2+a)} or 8 a ≤ 1 8a\leq {1} . Therefore a ≤ 1 8 a\leq {\dfrac{1}{8}} . Hence the maximum value of a a is 1 8 \boxed {\dfrac{1}{8}}

2 Helpful 0 Interesting 1 Brilliant 0 Confused
Patrick Corn
Dec 4, 2019

The derivative is 1 + a − x 2 ( a + x 2 ) 2 , 1 + \frac{a-x^2}{(a+x^2)^2}, and setting this equal to zero and solving leads to x 4 + ( 2 a − 1 ) x 2 + ( a 2 + a ) = 0 , x^4 + (2a-1)x^2 + (a^2+a) = 0, from which the quadratic formula gives x 2 = 1 − 2 a ± 1 − 8 a 2 . x^2 = \frac{1-2a \pm \sqrt{1-8a}}2. The largest value of a a that gives a nonnegative number under the square root is a = 1 8 . a = \frac18. This does in fact lead to valid solutions ( x 2 = 3 / 8 x^2 = 3/8 ), so the answer is 1/8 . \fbox{1/8}.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

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