Question

For any real number $x$, the maximum value of $4-6x-x^2$ is

• A. $13$
• B. $11$
• C. $9$
• D. $17$

Answer 1

The correct option is A $13$

$f\left(x\right)=4-6x-x^2$

For critical points,

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

$-6-2x=0$

$-2x=6$

$x=-3$

For maxima/minima

$f^{\prime \prime }\left(x\right)=-2$ (negative)

Hence at $x=-3f\left(x\right)$ will have maxima

$f(-3)=4-6\times (-3)-{(-3)}^2$

$=4+18-9$

$f(-3)=13$

Maximum value of $f\left(x\right)$ is $13$