Question

Find the local maximum and local minimum values $\frac{(x-1)(x-6)}{x-10},wherex\ne 10.$

Answer 1

$f\left(x\right)=\left(\frac{x^2-7x+6}{x-10}\right)f^{\prime }\left(x\right)=\left(\frac{(2x-7)(x-10)-(x^2-7x+6)\times 1}{(x-10)}\right)forcriticalpoints(2x-7)(x-10)-(x^2-7x+6)=02x^2-20x-7x+70-x^2+7x-6=0x^2-20x+64=0x^2-16x-4x+64=0x(x-16)-4(x-16)=0(x-4)(x-16)=0\therefore x=4,16\therefore localmaximaatx=4andf\left(4\right)=\left(\frac{3\times (-2)}{(-6)}\right)=1andlocalminimaatx=16andf\left(16\right)=\left(\frac{15\times 10}{6}\right)=25$