- #1
richyw
- 180
- 0
Homework Statement
I have the function
[tex]
h(x) =
\begin{cases}
x^2\sin{\left(\frac{1}{x}\right)}+\frac{x}{2} & \text{if } x \neq 0 \\
0 & \text{if } x =0
\end{cases}
[/tex]
and need to show that [itex]h'(0)>0[/itex] but there is no neighborhood of 0 on which h is increasing
Homework Equations
N/A
The Attempt at a Solution
well I can show that [itex]h'(0)=1/2>0[/itex] and if [itex]x\neq 0[/itex] then [tex]h'(x)=2x\sin{\left(\frac{1}{x}\right)}-\cos{\left(\frac{1}{x}\right)}+\frac{1}{2}[/tex] I'm a little lost on how I can show that there is no neighborhood of 0 on which h is increasing. This means that [itex]h'(x)>0[/itex] right? I think I'm a bit confused as to what this is asking me I guess.
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