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The temperature distribution in a metal rod given by the following function of the position $x \in \mathbb{R}$: $$T(x) = \frac{1 + 2x}{2 + x^2}$$

What is the maximal and minimal temperature in the metal rod?

$T'(x) = 0$ when $x = 1$ or $x = -2$. But I can't calculate the global maximum and global minimum because $T(x)$ does not belong to any intervals? Am I correct?

Thank you.

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  • $\begingroup$ Where is the given interval for that function? $\endgroup$ – Dr. Sonnhard Graubner Aug 11 '18 at 13:14
  • $\begingroup$ Try to use MathJax to format your question the next time. $\endgroup$ – mrtaurho Aug 11 '18 at 13:14
  • $\begingroup$ Thanks. Actually there are no any intervals in the question. That's why I am confused. $\endgroup$ – Mike LoongBoong Aug 11 '18 at 13:18
  • $\begingroup$ Try to draw this function with the help of derivative... $\endgroup$ – dmtri Aug 11 '18 at 13:31
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Many things become clearer when you graph the function.

enter image description here

\begin{align} \frac{d}{dx}\left(\frac{1 + 2x}{2 + x^2}\right) &= 0 \\ \dfrac{-2(x^2 + x - 2)}{(x^2+2)^2} &= 0 \\ x^2 + x - 2 &= 0 \\ (x+2)(x-1) &= 0 \\ x &\in \{-2, 1\} \end{align}

Note that $$\lim_{x \to \pm \infty} \frac{1 + 2x}{2 + x^2} = 0$$.

The max value is at $T(1)=1$ and the min value is at $T(-2)=-\dfrac 12.$

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Hint: I have got $$f_{max}=1$$ at $$x=1$$

$$f_{min}=-\frac{1}{2}$$ at $$x=-2$$

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Assuming your work on the derivative is correct (I haven't checked), since $T$ is continuous: \begin{align} T(1)&=1\\ T(-2)&=-\frac{1}{2}\\ T(-\infty)&=T(+\infty)=0 \end{align}

So there you have: global max at $x=1$ and global min at $x=-2$.

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  • $\begingroup$ Thanks a lot francescop21. It really helps. $\endgroup$ – Mike LoongBoong Aug 11 '18 at 13:52

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