David Young

2021-12-26

Q: Let $\{\begin{array}{cc}{x}^{m}\mathrm{sin}\frac{1}{x};& x\ne 0\\ 0;& x=0\end{array}$

Find the set of values of m for which

(i)$f\left(x\right)$ is continuous at $x=0$

(ii)$f\left(x\right)$ is differentiable at $x=0$

(iii)$f\left(x\right)$ is continuous but not differentiable at $x=0$ .

Find the set of values of m for which

(i)

(ii)

(iii)

Bernard Lacey

Beginner2021-12-27Added 30 answers

The given function is:

$\{\begin{array}{cc}{x}^{m}\mathrm{sin}\frac{1}{x};& x\ne 0\\ 0;& x=0\end{array}$

i) f is continuous at$x=0$ if

$f\left(0\right)=\underset{x\to 0}{lim}f\left(x\right)$

Now,$\underset{x\to 0}{lim}f\left(x\right)$

$=\underset{x\to 0}{lim}{x}^{m}\mathrm{sin}\frac{1}{x}$

$x=0$ is $m>0$

$\underset{x\to 0}{lim}f\left(0\right)\ne 0$ if $m<0$

$\delta}_{0$ f is continuous for $mt(0,\mathrm{\infty})$

i) f is continuous at

Now,

Jenny Bolton

Beginner2021-12-28Added 32 answers

ii) Now, ${f}^{\prime}\left(x\right)=m\cdot {x}^{m-1}\cdot \mathrm{sin}\frac{1}{x}+{x}^{m}\cdot \mathrm{cos}\left(\frac{1}{x}\right)\cdot \left(\frac{-1}{{x}^{2}}\right)$

$m{x}^{m-1}\cdot \mathrm{sin}\left(\frac{1}{x}\right)-{x}^{m-2}\mathrm{cos}\left(\frac{1}{x}\right)$ if $x\ne 0$

So, it is differentiable at $x=0$

if $m>2$ because then only

f(x) is differentiable

Hence, f is continuous but only differentiable at $x=0$ if $m\in [2,\mathrm{\infty})$

karton

Expert2022-01-09Added 613 answers

iii) So, f is continuous but only differentiable at

is on (0, 2)

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