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,
$f'(x ) = 12x^3- 24x^2 + 24x - 48$ and
$f''(x) = 36x^2 - 48x + 24$
Now, $f'(x) = 0$
$\Rightarrow 12x^3 - 24x^2 + 24x - 48 = 0$
$\Rightarrow x^3 - 2x^2 + 2x - 4 = 0$
$\Rightarrow x^2( x - 2) + 2 (x - 2) = 0$
$\Rightarrow (x-2)(x^2+2) = 0$
$\Rightarrow x = 2$ $[\because x^2 + 2 \ne 0]$
At $x = 2$, we have
$f''(x) = 36(2)^2 - 48(2) + 24 = 72 > 0$.
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.