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I had to minimise the function

$$f(x) = \frac{x^2 - x +4}{x-1}$$

I did the method where I found the range of this function and found the minimum value. However I know some basic calculus and was trying to find it using that but I am not able to. So, how do we find minima of this expression using calculus?

user0102
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Rasputin
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  • Were you able to differentiate the function wrt. $x$? Where did you get stuck? – Math Lover Oct 27 '20 at 09:57
  • Actually the minimum does not exist, since $$\lim_{x \to 1^-} f(x)= - \infty$$ – Crostul Oct 27 '20 at 10:16
  • @pH74 but the function is monotonically decreasing for $x \leq - 3$ and monotonically increasing for $x \geq 5$ and as Crostul said, it is undefined at $x = 1$. So there is no min or max. – Math Lover Oct 27 '20 at 10:34

2 Answers2

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HINT

First of all, notice that

\begin{align*} f(x) = \frac{x^{2} - x + 4}{x - 1} = \frac{x^{2} - x}{x-1} + \frac{4}{x-1} = x + \frac{4}{x-1} \end{align*}

Then determine for which values of $x$ one has that $f'(x) = 0$. After so, verify whether $f''(x) > 0$.

user0102
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  • What about functions which don’t neatly factor like this....Is there a differentiation method where we can simultaneously differentiate denominator and numerator? – Rasputin Oct 27 '20 at 11:45
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    Such result is known as the quotient rule for derivatives. More precisely, one has that

    $$\left(\frac{f(x)}{g(x)}\right)' = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^{2}}$$

    You can find out more about it at https://en.wikipedia.org/wiki/Quotient_rule

     – user0102 Oct 27 '20 at 11:47
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If $$f(x) = \frac{x^2 - x + 4}{x-1} = x + \frac{4}{x-1}$$ then $$f'(x) = 1 - \frac{4}{(x-1)^2} \text{ and }f''(x) = \frac{8}{(x-1)^3}$$ Can you take it from here?

stoic-santiago
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