# If the solve the problem

Question:

(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) $f(x)=x+\sqrt{1-x}, x \leq 1$

Solution:

(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

$f(-6)=(-6)^{4}-62(-6)^{2}+120 \times(-6)+9=-1647$