Question

Is Rolle's theorem applicable for the function $f\left(x\right)=x^3-6x^2+11x-6$?

Answer 1

Here, we have $f\left(x\right)=x^3-6x^2+11x-6,iff\left(x\right)=0$

Then $x^3-6x^2+11x-6x=0$
$\Rightarrow (x-1)(x-2)(x-3)=0$
$\therefore x=1,2,3.=f\left(1\right)=0=f\left(2\right)=f\left(3\right)$

Also $f^{{}^{\prime }}\left(x\right)=3x^2-12x+11$
Now $R{f}^{{}^{\prime }\left(x\right)}=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right))}{h}$

$=\underset{h\to 0}{lim}\frac{\left[\right(x+h{)}^3+6(x+h{)}^2+11(x+h)-6]-[x^3-6x^2+11x-6]}{h}$

$=\underset{h\to 0}{lim}\frac{(x+h{)}^3-x^3-6(x+h{)}^2-x^2+11(x+h)-x)}{h}$

$=\underset{h\to 0}{lim}\frac{(x+h{)}^3-x^3}{h}-6\underset{h\to 0}{lim}\frac{\left[\right(x+h{)}^2-x^2]}{h}+11\underset{h\to 0}{lim}\frac{(x+h)-x}{h}$

$=3x^2-12x+11.$
Similarly $L{f}^{{}^{\prime }}\left(x\right)=\underset{h\to 0}{lim}\frac{f(x-h)-f\left(x\right)}{-h}$

$=\underset{h\to 0}{lim}\frac{(x-h{)}^3-x^3}{-h}-6\underset{h\to 0}{lim}\frac{{(x-h)}^2-x^2}{-h}+11\underset{h\to 0}{lim}\frac{(x-h)-x}{-h}$

$=3x^2-12x+11$