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
Verified
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: \[f(x) = \left\{ \begin{gathered} (x - x)( - x){\text{ }},x < 0 \\ (x + x)x{\text{ }},x \geqslant 0 \\ \end{gathered} \right.\] \[f(x) = \left\{ \begin{gathered} 0{\text{ }},x < 0 \\ 2{x^2}{\text{ }},x \geqslant 0 \\ \end{gathered} \right.............(1)\] 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.
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