Find the maximum and minimum of the equation: $x^2-x+\frac{1}{x^2}+x+1$.
I am randomly trying to substitute values for $x$ but I need a complete method to solve such problems.
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The standard way to go about this would be through calculus. Local maxima and minima occur when the derivative is zero. wikipedia article – Samadin May 15 '17 at 17:09
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Canceling out the $-x$ and $+x$ terms could be a good starting point. – dxiv May 15 '17 at 18:08
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There must be some content lost when formatting, otherwise it would be just stupid to have $-x$ and $x$ in the same function. Was it supposed to be $\dfrac {x^2-x+1} {x^2+x+1}$ maybe? – Alex M. May 20 '17 at 17:23
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Downvoting because the OP doesn't even care to answer the questions that try to clarify the meaning of the (ambiguous) question, despite visiting the site (as seen from the profile page). – Alex M. May 21 '17 at 10:08
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This is a method without using calculus.
Let $f\colon \mathbb{R}\setminus\{0\}\to \mathbb{R}$, be the function defined by $f(x)=x^2+\frac{1}{x^2}+1$.
The function $f$ has no maximum because $x^2+\frac{1}{x^2}+1\geq x^2$ and $x^2$ can be as big as we want by getting $x$ bigger.
Consider then the following inequality: $$\left(x-\frac{1}{x}\right)^2\geq 0$$
This inequality is always true because every square of a real number is always nonnegative.
By expanding the left-hand side, we get: \begin{align*} x^2-2+\frac{1}{x^2}&\geq 0\\ x^2+\frac{1}{x^2}&\geq 2\\ x^2+\frac{1}{x^2}+1&\geq 3 \end{align*}
To finally prove that $3$ is the minimum, the points $1$ and $-1$ return $3$ when evaluating the function. So $f(x)\geq 3$ whenever $x\in \mathbb{R}\setminus\{0\}$ with equality when $x=\pm1$.
Tornado
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