A. $f(x)=3 x^4-2 x^3-6 x^2+6 x+1$
$f^{\prime}(x)=12 x^3-6 x^2-12 x+6$
$f^{\prime}(x)=6\left(2 x^3-x^2-2 x+1\right)$
$f^{\prime}(x)=6(x-1)(2 x-1)(x+1)$
Put, $\quad f^{\prime}(x)=0 x=-1, \frac{1}{2}, 1$


$f^{\prime}(x)$ is increasing in $\left(-1, \frac{1}{2}\right) \cup(1, \infty)$
decreasing value in $(\infty,-1) \cup(1 / 2, l)$
minimum value at $x=1$ or -1
maximum value of $x=\frac{1}{2}$
B. $f(x)=x+\frac{1}{x}$
$f^{\prime}(x)=1-\frac{1}{x^2}$ put $f^{\prime}(x)=0$
$\because \quad x^2=1 \Rightarrow x= \pm 1$
$f^{\prime \prime}(x)=\frac{2}{x^3}$
Maximum at $x=-1$
C. $\quad f(x)=x^4(7-x)^3$
$f^{\prime}(x)=4 x^3(7-x)^3-3 x^4(7-x)^2$
Put $\quad f^{\prime}(x)=0 \Rightarrow x^3(7-x)^2(28-7 x)=0$
$x=4$
Maximum at $x=4$
D. $\quad f(x)=x^4+(8-x)^4$
$f^{\prime}(x)=4 x^3-4(8-x)^3$, put $f^{\prime}(x)=0$
$x^3=(8-x)^3 \Rightarrow x=8-x$
$2 x=8 \Rightarrow x=4$
$f^{\prime \prime}(x)=12 x^2+12(8-x)^2$
$f^{\prime \prime}(4) \geq 0, \therefore$ Minimum of $x=4$