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Math Help - Showing absolute minimum

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    Showing absolute minimum

    Let f:\mathbb{R}\rightarrow\mathbb{R} be defined by f(x) = 2x^4+x^4\sin{1/x} for x\neq0 and f(0)=0 .
    Show that f has an absolute minimum at x=0, but that its derivative has both positive and negative values in every neighborhood of 0 .
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by frenchguy87 View Post
    Let f:\mathbb{R}\rightarrow\mathbb{R} be defined by f(x) = 2x^4+x^4\sin{1/x} for x\neq0 and f(0)=0 .
    Show that f has an absolute minimum at x=0, but that its derivative has both positive and negative values in every neighborhood of 0 .
    f(x)=2x^4+x^4\sin\left(\tfrac{1}{x}\right)\geqslan  t 2x^4-x^4=x^4\geqslant 0. Thus, f(x)\geqslant0,\text{ }\forall x\in\mathbb{R}. The fact that f(0)=0 let's you draw the proper conclusion.

    Now, f'(x)=\begin{cases} 8x^3+4x^3\sin\left(\tfrac{1}{x}\right)-x^2\cos\left(\tfrac{1}{x}\right) & \mbox{if} \quad x\ne 0 \\ 0 & \mbox{if}\quad x=0\end{cases}. Notice that f'(x)=x^2\left(8x+4x\sin\left(\tfrac{1}{x}\right)-\cos\left(\tfrac{1}{x}\right)\right) and so it suffices to show the claim for g(x)=8x+4x\sin\left(\tfrac{1}{x}\right)-\cos\left(\tfrac{1}{x}\right)


    Now, choose values of x small enough to fit into any range such that the first two terms are irrelevant.

    I leave this to you.
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