# Minimise the function $f(x) = \frac{x^2 - x +4}{x-1}$ using Calculus

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?

• Were you able to differentiate the function wrt. $x$? Where did you get stuck? Oct 27 '20 at 9:57
• Actually the minimum does not exist, since $$\lim_{x \to 1^-} f(x)= - \infty$$ 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. Oct 27 '20 at 10:34

\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 $$f''(x) > 0$$.
• 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 en.wikipedia.org/wiki/Quotient_rule Oct 27 '20 at 11:47
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?