Consider the function $f(t)=(max(0,t){)}^2$ for $-\mathrm{\infty }<t<\mathrm{\infty }$, where$max(a,b)$ denotes the maximum of a and b. Which of the following statements is/are true?

The correct statement is: f (t) is differentiable and its derivative is continuous.

To understand why this is correct, we can analyze the given function step-by-step by expressing it in a piecewise form and examining its differentiability and the continuity of its derivative.

Step 1: Express the function piecewise
The given function is:
$f(t)=(max(0,t){)}^2$
By definition, the maximum function behaves as follows:
$max(0,t)=0$ for $t\le 0$
$max(0,t)=t$ for $t>0$
Squaring these pieces gives us the piecewise representation of $f(t)$:
$f(t)=0$ when $t\le 0$
$f(t)=t^2$ when $t>0$

Step 2: Determine differentiability of $f(t)$
For intervals away from $t=0$, the derivative is straightforward:
If $t<0$, then $f^{\prime }(t)=0$.
If $t>0$, then ${\mathrm{f<mo>\prime </mo>}}^{}(t)=2t$.
Now we must check differentiability at the transition point $t=0$ by computing the left-hand derivative (LHD) and right-hand derivative (RHD):
Left-hand derivative at $t=0$:
$$\text{LHD}=\underset{h\to 0^{-}}{lim}\frac{f(0+h)-f(0)}{h}=\underset{h\to 0^{-}}{lim}\frac{0-0}{h}=0$$
Right-hand derivative at $t=0$:
$$\text{RHD}=\underset{h\to 0^{+}}{lim}\frac{f(0+h)-f(0)}{h}=\underset{h\to 0^{+}}{lim}\frac{h^2-0}{h}=\underset{h\to 0^{+}}{lim}h=0$$
Since $\text{LHD}=\text{RHD}=0$, the function $f(t)$ is differentiable at $t=0$, and its derivative is $f^{\prime }(0)=0$.

Step 3: Analyze the continuity of the derivative
The derivative function $f^{\prime }(t)$ can be written as:
$f^{\prime }(t)=0$ when $t\le 0$
$f^{\prime }(t)=2t$ when $t>0$
We examine the limit of $f^{\prime }(t)$ as $t$ approaches $0$:
Left-hand limit:
$$\underset{t\to 0^{-}}{lim}{f}^{\prime }(t)=\underset{t\to 0^{-}}{lim}0=0$$
Right-hand limit:
$$\underset{t\to 0^{+}}{lim}{f}^{\prime }(t)=\underset{t\to 0^{+}}{lim}2t=0$$
Since the limit of the derivative as $t\to 0$ equals $f^{\prime }(0)=0$, the derivative $f^{\prime }(t)$ is continuous everywhere.

Note that if we try to differentiate $f^{\prime }(t)$ once more, the derivative of $f^{\prime }(t)$ (which is $f^{″}(t)$) has a left-hand limit of $0$ and a right-hand limit of $2$ at $t=0$, meaning $f^{\prime }(t)$ is not differentiable at $t=0$.
Therefore, $f(t)$ is differentiable and its derivative is continuous, but the derivative is not itself differentiable.