Question

In the mean value theorem, $f'(c) = \frac{f(b)-f(a)}{b-a}$, if $\text{a} = 0$, $\text{b} = \frac{1}{2}$ and $f(x) = x(x - 1)(x - 2)$, then the value of $c$ is

• A. $1 - \frac{\sqrt{15}}{6}$
• B. $1 - \frac{\sqrt{13}}{6}$
• C. $1 - \frac{\sqrt{21}}{6}$
• D. $1 + \frac{\sqrt{21}}{6}$

Hint:

Choose value of $c$ within the interval.

Answer 1

The Correct Option is B

Step 1: Compute RHS. \[ f(0)=0,\quad f\left(\frac{1}{2}\right)=\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)=\frac{3}{8} \] \[ \frac{f(b)-f(a)}{b-a} = \frac{\frac{3}{8}}{\frac{1}{2}} = \frac{3}{4} \]
Step 2: Find derivative. \[ f(x)=x^3 - 3x^2 + 2x \] \[ f'(x)=3x^2 - 6x + 2 \]
Step 3: Apply MVT. \[ 3c^2 - 6c + 2 = \frac{3}{4} \] \[ 12c^2 - 24c + 8 = 3 \] \[ 12c^2 - 24c + 5 = 0 \]
Step 4: Solve. \[ c = \frac{24 \pm \sqrt{576 - 240}}{24} = \frac{24 \pm \sqrt{336}}{24} \] \[ = 1 \pm \frac{\sqrt{21}}{6} \] Only value in $(0,\frac{1}{2})$: \[ c = 1 - \frac{\sqrt{13}}{6} \]
Step 5: Conclusion. \[ {1 - \frac{\sqrt{13}}{6}} \]