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## Homework Statement

[/B]

Let ##f:ℝ→ℝ## such that ##f(x)= \left\{

\begin{array}{l}

2x^4+x^4sin(\frac{1}{x}), x≠0 \\

0 , x=0 \\

\end{array}

\right. ##

a.) Prove ##f## is continuous and differentiable everywhere. Find the (global) minimum (##f(x_0)##) of ##f## and point ##x_0##.

b.) Prove there doesn't exist half-open interval ##[x_0,b[##, ##b>x_0##, where ##f'(x)≥0## ##∀x∈[x_0,b[##.

## The Attempt at a Solution

a.) When ##x≠0##, ##f## is continuous as a sum of continuous functions and ##f'(x)=8x^3+4x^3sin(1/x)-x^2cos(1/x)##.

When ##x=0##, ##f## is continuous because ##\lim_{x \rightarrow 0} 2x^4+x^4sin(\frac{1}{x})=...=0=f(0)## and ##\lim_{x \rightarrow 0} \frac{2x^4+x^4sin (1/x)-0}{x-0}=\lim_{x \rightarrow 0} 2x^3+\lim_{x \rightarrow 0}x^3sin (1/x)=0+0=0=f'(0)##.

My idea is to show that ##f(x)>0## if ##x∈]0, π/2[ ##and ##f(x)<0## if ##x∈]-π/2, 0[##. Then 0 would be local and therefore global minimum?

b.) no idea

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