Will you integrate it 2015 times? - X

Calculus Level 4

A function $f : \mathbb{R} \rightarrow \mathbb{R}$ is differentiated $2015$ times to get $g(x) = \dfrac{ d^{2015} }{dx^{2015} } \big[ f(x) \big]$ .

If $g(x)$ is a non-constant function defined for all real values of $x$ , then what is the minimum number of distinct solutions $f(x) = g(x)$ can possess?

This problem is part of the set - Will you integrate it 2015 times? .

2015 2013 1 2014 0 None of the these choices. 2 2016

1 solution

Pranshu Gaba
Nov 12, 2015

Consider the function $f(x) = e^{-x}$ for all $x$ . After differentiating it $2015$ times, we get $g(x) = -e^{-x }$ for all $x$ .

Note that $f( x ) = g(x)$ means $e^{-x} = - e ^{-x} \implies 2 e^{-x } = 0$ .

This has zero solutions since $e^{-x}$ is always greater than $0$ . Since it is not possible to have less than $0$ solutions, the minimum number of solutions of $f(x) = g(x)$ are $\boxed{0}$ $_\square$

Will you integrate it 2015 times? - X

This problem is part of the set - Will you integrate it 2015 times? .

1 solution

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