Q. 14 - Calculus

Question

Find the critical points of the function  

$f^{\text{'}}\left(x\right)=\frac{(x-1)(x-2)}{x-3}$

Step-by-Step Solution

Answer

$T\mathrm{he} \mathrm{critical} \mathrm{points} \mathrm{of} \mathrm{the} \mathrm{function} \mathrm{is} x = 1, 2$

Step 1. Given

The given function is $f^{\text{'}}\left(x\right)=\frac{(x-1)(x-2)}{x-3}$.

Step 2. Calculation

$$\begin{gathered} "\mathrm{A} \mathrm{point} \mathrm{x} = \mathrm{c} \mathrm{in} \mathrm{the} \mathrm{domain} \mathrm{of} \mathrm{f} \mathrm{is} \mathrm{called} \mathrm{a} \mathrm{critical} \mathrm{point} \mathrm{of} \mathrm{fif} \mathrm{f}\text{'}\left(\mathrm{c}\right)=0 \mathrm{or} \mathrm{f}\text{'}\left(\mathrm{c}\right)\mathrm{does} \mathrm{not} \mathrm{exist}. \\ Now, equate f\text{'}\left(x\right)=0 \\ \frac{ \left(\right(x - 1\left)\right(x - 2\left)\right)}{(x - 3) }= 0 \\ (x - 1)(x - 2) = 0 \\ x = 1, x = 2 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{critical} \mathrm{points} \mathrm{of} \mathrm{the} \mathrm{function} \mathrm{is} x = 1, 2 \end{gathered}$$