Continuity And Differentiability

[himanshu121]

Consider [tex]f(x)=x^3-x^2+x+1 [/tex]
[tex] g(x)=\left\{\begin{array}{cc}{max\{f(t),0\leq t \leq x\}}\;\ 0\leq x \leq 1
\\ 3-x\;\ 1< x \leq 2\end{array}\right[/tex]

Discuss the continuity and differentiability of g(x) in the interval (0,2)

I know how to do it
As f(x) is increasing function therefore max will be x^3-x^2+x+1.
But
I want to know the problem graphically ??[?]

 

[Moni]

Oh! I am also in difficulty with "Continuity And Differentiability" like limits! Seems you can explain a little :)

 

[M98Ranger]

The graph of f(x)=x^3-x^2+x+1 is a cubic function with a positive leading coefficient, meaning it is an upward facing parabola. The graph is continuous and differentiable for all values of x.

The graph of g(x) is a piecewise function, with the first piece being the same as f(x) and the second piece being a straight line with a negative slope.

In the interval (0,1), g(x) will be equal to f(x) and will have the same continuity and differentiability as f(x).

In the interval (1,2), g(x) will be equal to 3-x, which is a continuous and differentiable function.

Therefore, g(x) is continuous and differentiable in the interval (0,2) as it is made up of continuous and differentiable functions in each subinterval.

Graphically, this can be seen as a smooth curve for the first part of the graph (0,1) and a straight line for the second part (1,2). The graph will have no breaks or sharp turns, indicating continuity, and it will have a smooth slope throughout, indicating differentiability.