Question

Draw the graph of $y=12x^3-4x^2-3x+1$. Hence find the number of positive zeroes.

Answer 1

REF.Image.

$y=f\left(n\right)=12x^3-4x^2-3x+1$

$f^{\prime }\left(n\right)=36x^2-3x-3=0$

$\Rightarrow x=\frac{8\pm \sqrt{64+432}}{72}$

$=\frac{8\pm \sqrt{496}}{72}$

$8\pm 4\sqrt{31}/72$

$f\left(0\right)=1$

$f\left(1/2\right)=12/8-4/4-3/2+1$

$=-7/8<0$

$f\left(1\right)=12-4-3+1>0$

$\therefore$ No of $+ve$ roots $=2$

$x_1=\frac{8-4\sqrt{31}}{72}$

$x_2=\frac{8+4\sqrt{31}}{72}$