Functions#14

Calculus Level 3

2 3 1 4

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2 solutions

Marta Reece
Feb 26, 2018

The function goes smoothly from limit in minus infinity of 1 2 \frac12 to a zero at x = 0 x=0 . It remain zero until x = 1 x=1 . It rises with slope 1 to the value of 1 at x = 2 x=2 . It remains at 1 until x = 3 x=3 . From the point ( 3 , 1 ) (3, 1) , it drops smoothly to its limit of 1 2 \frac12 at infinity.

The graph of the function is symmetrical around a point ( 1.5 , 0.5 ) (1.5, 0.5) . However the function is not odd, as this point is not ( 0 , 0 ) (0, 0) .

It is clearly not a bijection or one-one function.

It does, however, realize every x x on the interval [ 0 , 1 ] [0,1] , so it is onto.

Aakhyat Singh
Feb 26, 2018

@Marta Reece , how did you do this question ?

I found the function very interesting. I posted my solution, such as it is. I hope that answers your question.

Marta Reece - 3 years, 3 months ago

I wonder why you asked me this. Didn't you know how to solve a problem you posted?

Marta Reece - 3 years, 2 months ago

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