real world applications - derivatives

[tmd1979]

I am using a postage function, where C(x)= .37 if 0<x<=1, .60 if 1<x<=2, .83 if 2<x<=3, 1.06 if 3<x<=4, and 1.29 if 4<x<=5. I am supposed to list intervals where the function has derivative and include what is this derivative.

If I am not mistaken, the funjction has a derivative where x is not an integer, right? Would that be (0,1), (1,2), (2,3), (3,4), (4,5]? Wouldn't the derivative be 0 because isn't that the derivative of a number without a variable?

I am a little confused as to what they are asking me to accomplish.

 

[soroban]

Hello, tmd1979!

\(\displaystyle \text{I am using a postage function, where: }\; C(x)\;=\;\left\{\begin{array}{cccc}0.37 && 0<x\le 1 \\0.60 && 1<x\le2 \\ 0.83 && 2<x \le3 \\ 1.06 && 3<x\le4 \\ 1.29 && 4<x\le5 \end{array}\right\)

\(\displaystyle \text{I am supposed to list intervals where the function has derivative, and include what this derivative is.}\)

\(\displaystyle \text{If I am not mistaken, the function has a derivative where }x\text{ is not an integer, right?}\) . Right!

\(\displaystyle \text{Would that be: }\; (0,1),\;(1,2),\;(2,3),\;(3,4),\;(4,5]\:?\)
I believe that "5" is not included.
The function is not differentiable at x = 5.
The "two-sided limit" does not exist.

\(\displaystyle \text{Wouldn't the derivative be 0, because that's the derivative of a constant?}\) . Yes!

As we say in the Corps, "Ya done good!"

 

[tmd1979]

Thanks for reaffirming my thoughts. Your help is greatly appreciated.

Is this same function integrable? Can you explain why or why not?

Can the definite integral, C(x), be calculated over closed intervals that contain the points of discontinuity. Please explain of how this might be done.

One more thing, for example, what would be an integral of C(x) over interval [0.5, 2] and how this would be calculated?

I am thinking [.60(2)-.60(1)] +[.37(1)-.37(.5)]=.6+.185=.785

I don't think I am even close.