Question

If $2\le x\le 4,$ then the maximum value of
$f\left(x\right)=(x-2{)}^6(4-x{)}^5$ is

• A. ${\left(\frac{2}{11}\right)}^6{\left(\frac{10}{11}\right)}^5$
• B. ${\left(\frac{12}{11}\right)}^6{\left(\frac{10}{11}\right)}^5$
• C. ${\left(\frac{11}{12}\right)}^6{\left(\frac{11}{10}\right)}^5$
• D. ${\left(\frac{2}{11}\right)}^6{\left(\frac{10}{9}\right)}^5$

Answer 1

The correct option is B ${\left(\frac{12}{11}\right)}^6{\left(\frac{10}{11}\right)}^5$

$f\left(x\right)=(x-2{)}^6(4-x{)}^5,$
$\Rightarrow f^{\prime }\left(x\right)=6(x-2{)}^5(4-x{)}^5-5(x-2{)}^6(4-x{)}^4(-1)=(x-2{)}^5(4-x{)}^4\left[6\right(4-x)-5(x-2\left)\right]=(x-2{)}^5(4-x{)}^4(34-11x)$
Now, $f^{\prime }\left(x\right)=0$
$\Rightarrow x=2,\frac{34}{11},4$
$\Rightarrow f\left(2\right)=f\left(4\right)=0$
$\Rightarrow f\left(\frac{34}{11}\right)={\left(\frac{12}{11}\right)}^6{\left(\frac{10}{11}\right)}^5,$ which is the maximum value.