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  • Let  be defined as follows: <img alt="\begin{array}{r} \mathrm{f(x)=x^6, x^2>1

Let \text{f(x)} be defined as follows:

\begin{array}{r} \mathrm{f(x)=x^6, x^2>1 }\\ \mathrm{x^3, x^2 \leq 1 .} \end{array}

Then \mathrm{f(x)} is

Option: 1

 continuous everywhere
 


Option: 2

 differentiable everywhere
 


Option: 3

 discontinuous at x=-2
 


Option: 4

 not differentiable at x=1


Answers (1)

best_answer

\begin{aligned} \mathrm{x^2>1 \Rightarrow }&\mathrm{ x<-1 \text { or } x>1 ; \quad x^2 \leq 1 \Rightarrow-1 \leq x \leq 1 . }\\ \therefore \quad\mathrm{ f(x)}= & \mathrm{x^6, x<-1} \\ &\mathrm{ x^3,-1 \leq x \leq 1} \\ & \mathrm{x^6, x>1} . \end{aligned}

Use these definitions. The only doubtful points for continuity and differentiability are \mathrm{x=-1,1}.

Clearly, \mathrm{f(-1-0)=1} and \mathrm{f(-1+0)=-1}.

So, \mathrm{f(x)} is not continuous at \mathrm{x=-1}.

Also \mathrm{f(1-0)=1, f(1+0)=1}  and \mathrm{f(1)=1}. So, \mathrm{f(x)}  is continuous at \mathrm{x=1}.

As \mathrm{f(x)} is not continuous at \mathrm{x=-1}, it is not differentiable there.

Now, \mathrm{ \lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}=\lim _{h \rightarrow 0} \frac{(1+h)^6-1}{(1+h)-1}=6 \cdot 1^{6-1}=6.}

\mathrm{ \lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}=\lim _{h \rightarrow 0} \frac{(1-h)^3-1}{(1-h)-1}=3 \cdot 1^{3-1}=3 .}

\mathrm{\therefore f(x)} is not differentiable at  \mathrm{x=1}.

Posted by

Ritika Harsh

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