A calculus problem by Akshat Sharda

Calculus Level 4

$f(x)=\left \lfloor x-\dfrac{1}{4} \right \rfloor + x \lfloor x \rfloor + | x(x-4) \sin x |+(2x-1)^{\frac{1}{3}}$

Find the number of points for $x\in [0,2\pi)$ at which $f(x)$ is non-differentiable.

Notations:

$\lfloor \cdot \rfloor$ denotes the floor function .
$| \cdot |$ denotes the absolute value function .

The answer is 15.

2 solutions

Indraneel Mukhopadhyaya
Feb 24, 2017

x={0.25,1.25,2.25,...6.25} U {1,2,3,...6} U {4,π} U {0.5} are the points of non - differentiability.So, there are 15 total points (4 is there 2 times).

Karan Kapoor
Feb 24, 2017

Well to me there are 14 points, until you count one two times.

There are 15 points. 7 points for [x-0.25], where it becomes an integer in [0,2π) ; 6 points for x[x] where it becomes an integer in [0,2π); (0 is not valid).And 2 points 4,π for the function |x(x-4)(sin(x))| , again 0 is not valid. And finally x=1/2 for (2x-1)^(1/3) where it has a vertical tangent.So, there are 15 points (4 is counted twice for x[x] and |x(x-4)(sin(x))| ).

Indraneel Mukhopadhyaya - 4 years, 3 months ago

A calculus problem by Akshat Sharda

The answer is 15.

2 solutions

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