Question

Find both the maximum and minimum values respectively of $3x^4-8x^3+12x^2-48x+1$ on the interval $[1,4]$.

• A. $-63$, $257$
• B. $257$, $-40$
• C. $257$, $-63$
• D. $63$, $-257$

Answer 1

Answer: (C)

Let $f(x)=3x^4-8x^3+12x^2-48x+1$. Then,
$f^{\prime }(x)=12x^3-24x^2+24x-48$ and
$f^{\prime \prime }(x)=36x^2-48x+24$
Now, $f^{\prime }(x)=0$
$\Rightarrow 12x^3-24x^2+24x-48=0$
$\Rightarrow x^3-2x^2+2x-4=0$
$\Rightarrow x^2(x-2)+2(x-2)=0$
$\Rightarrow (x-2)(x^2+2)=0$
$\Rightarrow x=2$ $[\because x^2+2\ne 0]$
At $x=2$, we have
$f^{\prime \prime }(x)=36(2{)}^2-48(2)+24=72>0$.
So, $x=2$ is a point of local minimum.
Now, $f(2)=-63$, $f(1)=-40$ and $f(4)=257$.
So, the minimum and maximum values of $f(x)$ on $[1,4]$ are $-63$ and $257$ respectively.