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.

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

Many things become clearer when you graph the function.

\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 $T(-2)=-\dfrac 12.$

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

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

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 $x=-2$.

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