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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
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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.
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.
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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