Question

The absolute maximum and minimum value of $f\left(x\right)=3x^4-8x^3+12x^2-48x+25,x\in [0,3]$ are respectively

• A. $-25,39$
• B. $25,-39$
• C. $8,-8$
• D. $8,10$

Answer 1

The correct option is B $25,-39$

Differentiate the function $f\left(x\right)=3x^4-8x^3+12x^2-48x+25$ with respect to $x$,

$f^{\prime }\left(x\right)=12x^3-24x^2+24x-48$

Put $f^{\prime }\left(x\right)=0$,

$12x^3-24x^2+24x-48=0$

$\left(x-2\right)\left(12x^2+24\right)=0$

$\left(x-2\right)12\left(x^2+2\right)=0$

$x-2=0$

$x=2$

Or,

$x^2=-2$

$x=\sqrt{2}i$

But $\sqrt{2}i$ is an imaginary number, so the value to be considered is $x=2$.

Now, substitute the value of $x$ and the end points in the given function.

$f\left(0\right)=3{\left(0\right)}^4-8{\left(0\right)}^3+12{\left(0\right)}^2-48\left(0\right)+25$

$=25$

$f\left(2\right)=3{\left(2\right)}^4-8{\left(2\right)}^3+12{\left(2\right)}^2-48\left(2\right)+25$

$=-39$

$f\left(3\right)=3{\left(3\right)}^4-8{\left(3\right)}^3+12{\left(3\right)}^2-48\left(3\right)+25$

$=10$

Therefore, the maximum value is $25$ and the minimum value is $-39$.