Question

# Let $f(x) = (x + \left| x \right|)\left| x \right|$. Which of the following is true for all x.(a) f is continuous(b) f is differentiable for some x(c) f’ is continuous(d) f’’ is continuous

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Hint: Write the expression of f(x) for x>0 and x<0. Check if f(x) is continuous, if it is continuous check for its differentiability. Then check if f’ and f’’ are continuous.

We know that $\left| x \right|$ is equal to x for x $\geqslant$ 0 and -x for x < 0. Then, we can express f(x) as follows:
$f(x) = \left\{ \begin{gathered} (x - x)( - x){\text{ }},x < 0 \\ (x + x)x{\text{ }},x \geqslant 0 \\ \end{gathered} \right.$
$f(x) = \left\{ \begin{gathered} 0{\text{ }},x < 0 \\ 2{x^2}{\text{ }},x \geqslant 0 \\ \end{gathered} \right.............(1)$
We know that constant and polynomial functions are continuous in their domain, so f(x) is continuous for all x > 0 and x < 0. Let us check the continuity of f(x) at x = 0.
The left-hand limit of f(x) at x=0 is as follows:
$LHL = \mathop {\lim }\limits_{x \to {0^ - }} f(x)$
$LHL = \mathop {\lim }\limits_{x \to 0} 0$
$LHL = 0............(2)$
Hence, the LHL of f(x) at x = 0 is zero.
Now, the right-hand limit of f(x) at x = 0 is as follows:
$RHL = \mathop {\lim }\limits_{x \to {0^ + }} f(x)$
$RHL = \mathop {\lim }\limits_{x \to 0} 2{x^2}$
$RHL = 0............(3)$ The value of f(x) at x = 0 is as follows:
$f(0) = 2{(0)^2}$
$f(0) = 0..........(4)$
From equations (2), (3) and (4), we have:
$LHL = RHL = f(0)$
Hence, f(x) is continuous at x = 0.
Therefore, f(x) is continuous everywhere.
We now check the differentiability of f(x) at x=0.
The left-hand derivative of f(x) at x = 0 is as follows:
$LHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x - h) - f(x)}}{{ - h}}$
Here, x = 0, then, we have:
$LHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{f( - h) - f(0)}}{{ - h}}$
Using equation (1) in the above equation, we have:
$LHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{0 - 2{{(0)}^2}}}{{ - h}}$
$LHD = \mathop {\lim }\limits_{h \to 0} 0$
$LHD = 0..........(5)$
The right-hand derivative of f(x) is given as follows:
$RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$
Here, x = 0, then, we have:
$RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(h) - f(0)}}{h}$
Using equation (1), in the above equation, we obtain:
$RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{2{h^2} - 0}}{h}$
Simplifying further, we get:
$RHD = \mathop {\lim }\limits_{h \to 0} 2h$
$RHD = 0.........(6)$
From equation (5) and equation (6), we have:
$LHD = RHD$
Hence, f(x) is differentiable everywhere.
We find the derivative of f(x) as follows:
$f'(x) = \left\{ \begin{gathered} 0{\text{ }},x < 0 \\ 4x{\text{ }},x \geqslant 0 \\ \end{gathered} \right.............(7)$
Again, this is continuous for x > 0 and x < 0.
At x = 0, we have:
$LHL = \mathop {\lim }\limits_{x \to {0^ - }} f'(x)$
$LHL = \mathop {\lim }\limits_{x \to {0^ - }} 0$
$LHL = 0$
$RHL = \mathop {\lim }\limits_{x \to {0^ + }} f'(x)$
$RHL = \mathop {\lim }\limits_{x \to 0} 4x$
$RHL = 0$
$f'(0) = 0$
Hence, we have:
$LHL = RHL = f'(0)$
Therefore, f’(x) is continuous everywhere.
We now find f’’(x).
$f'(x) = \left\{ \begin{gathered} 0{\text{ }},x < 0 \\ 4{\text{ }},x \geqslant 0 \\ \end{gathered} \right.............(8)$
We know clearly that at x = 0,
$LHL = 0$
$RHL = 4$
$LHL \ne RHL$
Hence, f’’(x) is not continuous.
Hence, the correct options are (a), (b) and (c).

Note: To check continuity, finding left-hand limit and right-hand limit and equating them is not sufficient, we need to also check the value of the function at that point. A function is differentiable only if the function is continuous.