Question

The maximum value of $f\left(x\right)=x^3-9x^2+24x+5$ in the interval $[1,6]$ is

• A. $21$
• B. $25$
• C. $41$
• D. $46$

Answer 1

The correct option is C $41$

$f\left(x\right)=x^3-9x^2+24x+5$
$f^{\prime }\left(x\right)=3x^2-18x+24$
For maxima or minima:
$f^{\prime }\left(x\right)=0$
$\Rightarrow x^2-6x+8=0$
$\Rightarrow x=2,4$

$f\left(2\right)=(2{)}^3-9(2{)}^2+24\left(2\right)+5=25$
$f\left(4\right)=(4{)}^3-9(4{)}^2+24\left(4\right)+5=21$
Values at $f\left(x\right)$ boundaries:
$f\left(1\right)=(1{)}^3-9(1{)}^2+24\left(1\right)+5=21$
$f\left(6\right)=(6{)}^3-9(6{)}^2+24\left(6\right)+5=41$
The maximum value of $f\left(x\right)$ in interval $[1,6]$ is $41$.