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

Answer 1

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.

Complete step-by-step answer:
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:

(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.