Question

The function $f(x)=x^{3}-4 x^{2}+4 x+3$ defined on $[-1,3]$ has

• A. minimum value -6 at $x=-1$
• B. minimum value 6 at $x=3$
• C. minimum value 3 at $x=2$
• D. maximum value 9 at $x=3$

Answer 1

Answer: (A)

We have
$f(x)=x^{3}-4 x^{2}+4 x+3$
$\therefore f'(x)=3 x^{2}-8 x+4$
Now, $f'(x)=0$
$ \Rightarrow 3 x^{2}-8 x+4=0$
$\Rightarrow 3 x^{2}-6 x-2 x+4=0$
$\Rightarrow 3 x(x-2)-2(x-2)=0$
$\Rightarrow (x-2)(3 x-2)=0 $
$\Rightarrow x=2, \frac{2}{3}$
Now, $f(-1)=-1-4-4+3=-6$
$f(2 / 3)=\frac{8}{27}-4 \times \frac{4}{9}+4 \times \frac{2}{3}+3$
$=\frac{8-48+72+81}{27}=\frac{113}{27}$
$f(2)=8-14+8+3=5$
$f(3)=27-36+12+3=6$
$\therefore f(x)$ has minimum value at $x=-1$