Help Solving Two Math Problems - Maths Freak in Need

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well hello
i have these two questions which is giving a lot of problem can some one help me out.

i have this function
y = min(|x|,|x-1|,|x+1|)
how do we find point where this function is not differentiable and continous.
how dow we evaluate min(...)

secondly
[x+1.5]= [x +[sinx + [x + cosx]]]
how do we solve this [] represents greatest integer function.
x is and Integer.

thank you
Maths freak
:frown:
 
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Write y in a form that is easier to interpret.
Notice that:
y=|x+1|, for x <= -0.5
y=|x|, for -0.5 < x < 0.5
y=|x-1|, for x >= 0.5

You can graph this function for a better picture.
 
Hi Maths Freak,

I'm happy to help you with these math problems! Let's take a look at the first one:

To find the points where the function y = min(|x|, |x-1|, |x+1|) is not differentiable and continuous, we need to first understand what these terms mean.

A function is differentiable if it has a well-defined derivative at every point. This means that the function is smooth and has a tangent line at every point. On the other hand, a function is continuous if it has no sudden jumps or breaks in its graph. This means that the function is defined and has a limit at every point.

In order for a function to be differentiable, it must also be continuous. However, the reverse is not always true. A function can be continuous without being differentiable.

Now, let's look at our function y = min(|x|, |x-1|, |x+1|). This function is made up of three different absolute value functions. We can see that at x = 0, x = 1, and x = -1, there are points where two of the absolute value functions are equal. This means that at these points, the function is not differentiable because it has sharp corners or "cusps."

To find the points where the function is not continuous, we need to find the points where the minimum value of the three absolute value functions changes. This occurs at x = 0, x = 1, and x = -1. At these points, the function has a sudden jump or break in its graph, making it not continuous.

As for evaluating the min(...) part of the function, we can do this by comparing the values of the three absolute value functions at a given x-value and choosing the smallest value. For example, if we want to evaluate min(|1|, |1-1|, |1+1|), we would compare the values of 1, 0, and 2 and choose the smallest, which is 0.

Moving on to the second problem:

[x+1.5] = [x + [sinx + [x + cosx]]]

Here, we have a function that involves the greatest integer function, denoted by []. This function takes a number and rounds it down to the nearest integer. For example, [2.7] =
 

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