(i) $f(x)=x^{4}-62 x^{2}+120 x+9$
(ii) $f(x)=x^{3}-6 x^{2}+9 x+15$
(iii) $f(x)=(x-1)(x+2)^{2}$
(iv) $f(x)=2 / x-2 / x^{2}, x>0$
(v) $f(x)=x e^{x}$
(vi) $f(x)=x / 2+2 / x, x>0$
(vii) $f(x)=(x+1)(x+2)^{1 / 3}, x \geq-2$
(viii) $f(x)=x \sqrt{32-x^{2}}, \quad-5 \leq x \leq 5$
(ix) $f(x)=x^{3}-2 a x^{2}+a^{2} x, a>0, x \in R$
(x) $\mathrm{f}(\mathrm{x})=x+\frac{a 2}{x}, a>0, \mathrm{x} \neq 0$
(xi) $f(x)=x \sqrt{2-x^{2}}-\sqrt{2} \leq x \leq \sqrt{2}$
(xii) $f(x)=x+\sqrt{1-x}, x \leq 1$
(i)
Given : $f(x)=x^{4}-62 x^{2}+120 x+9$
$\Rightarrow f^{\prime}(x)=4 x^{3}-124 x+120$
For the local maxima or minima, we must have
$f^{\prime}(x)=0$
$\Rightarrow 4 x^{3}-124 x+120=0$
$\Rightarrow x^{3}-31 x+30=0$
$\Rightarrow(x-1)\left(x^{2}+x-30\right)=0$
$\Rightarrow(x-1)(x+6)(x-5)=0$
$\Rightarrow x=1,5$ and $-6$
Thus, $x=1, x=5$ and $x=-6$ are the possible points of local maxima or local minima.
Now,
$f^{\prime \prime}(x)=12 x^{2}-124$
At $x=1:$
$f^{\prime \prime}(1)=12(1)^{2}-124=-112<0$
So, $x=1$ is the point of local maximum.
The local maximum value is given by
$f(1)=1^{4}-62(1)^{2}+120 \times 1+9=68$
At $x=5:$
$f^{\prime \prime}(5)=12(5)^{2}-124=176>0$
So, $x=5$ is the point of local minimum.
The local minimum value is given by
$f(5)=5^{4}-62(5)^{2}+120 \times 5+9=-316$
At $x=-6:$
$f^{\prime \prime}(-6)=12(-6)^{2}-124=308>0$
So, $x=-6$ is the point of local maximum.
The local minimum value is given by
$f(-6)=(-6)^{4}-62(-6)^{2}+120 \times(-6)+9=-1647$
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