Question

The function $f(x)=x^3-4x^2+4x+3$ 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
$f(x)=x^3-4x^2+4x+3$
$\therefore f^{\prime }(x)=3x^2-8x+4$
Now, $f^{\prime }(x)=0$
$\Rightarrow 3x^2-8x+4=0$
$\Rightarrow 3x^2-6x-2x+4=0$
$\Rightarrow 3x(x-2)-2(x-2)=0$
$\Rightarrow (x-2)(3x-2)=0$
$\Rightarrow x=2,\frac{2}{3}$
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 $x=-1$