Critically Acclaimed Point

Calculus Level 3

Let f ( x ) = 1 1 + x 4 + A f(x)=\dfrac{1}{1+x^4}+A , where A A is a constant, and let F ( x ) F(x) be an antiderivative of f f . Find the value of A A for which F F has exactly one critical point.

Source: Clemson Calculus Challenge's Sample Problems


The answer is -1.

Adam Hufstetler by Adam Hufstetler , 20, USA

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1 solution

Adam Hufstetler
Mar 21, 2017

f ( x ) = 1 1 + x 4 f(x)=\frac{1}{1+x^4}
f ′ ( x ) = − 4 x 3 ( 1 + x 4 ) 2 f'(x)=\frac{-4x^3}{(1+x^4)^2}
0 = − 4 x 3 ( 1 + x 4 ) 2 0=\frac{-4x^3}{(1+x^4)^2}
x = 0 , ± ∞ x=0,\pm\infty
f f has an absolute maximum at x = 0 x=0 and absolute minimums at ± ∞ \pm\infty as these are the only relative extremum, with x = 0 x=0 being the only relative maximum and x = ± ∞ x=\pm\infty converging to the same value. In order to have one critical point on F F , f f 's absolute maximum must be shifted down or up to zero.
f ( 0 ) = 1 + A f(0)=1+A
A = − 1 A=-1


1 Helpful 0 Interesting 0 Brilliant 0 Confused

Oh, never mind. I misread.

The solution could use some clarity of explaining what you're doing, and why you're taking those steps. IE "We want F to have exactly one critical point, which means that f(x) = 0 has a unique solution. So, we ... "

Are you sure? We want the antiderivative, not the derivative.

Calvin Lin Staff - 4 years, 1 month ago

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