(1)

Step 1:

The function is $\"\\\\f\\left$

Let $\"\\\\f\\left$ .

$\"y=e^{x}+e^{-x}\\rightarrow\\left$

$\"\\\\f\\left$

Differentiate with respective x.

$\"\\\\f\'\\left$

Sum rule of derivatives : $\"\"$ .

$\"f\'\\left$

$\"f\'\\left$

$\"f\'\\left$ .

Step 2:

To find the critical numbers of $\"f\\left$ , equate $\"f\'\\left$ to zero.

$\"f\'\\left$ .

$\"e^{x}$ .

$\"e^{x}$

$\"\\\\e^{2x}-{1}$

Apply logrithm on each side.

$\"\\\\\\ln$

Substitute $\"x=0\"$ in equation (1).

$\"\\\\y=e^{0}+e^{\\left$

$\"y=2\"$ .

Critical values are : $\"x=0\"$ and $\"y=2\"$ .

Solution :

Critical values are : $\"x=0\"$ and $\"y=2\"$ .

(2)

Step 1:

The function is $\"\\\\g\\left$

Let $\"\\\\g\\left$ .

$\"\\\\y=x^{3}\\ln$

$\"\\\\g\\left$

Differentiate with respective x.

$\"\\\\g\'\\left$

Product rule of derivatives : $\"\\frac{d}{dx}\\left$ .

$\"g\'\\left$

$\"g\'\\left$

$\"\\\\g\'\\left$

Step 2:

To find the critical numbers of $\"g\\left$ , equate $\"g\'\\left$ to zero.

$\"x^{2}\\left$

$\"x^{2}=0\"$ and $\"(1+3\\ln$

since $\"x=0\"$ is not in the domain so $\"x=0\"$ is not consider.

$\"\\ln$

$\"x$

$\"x=0.71\"$

Substitute $\"x=0\"$ in equation (1).

$\"\\\\y=\\left$

$\"y=-0.122.\"$

Critical values are : $\"x=0.71\"$ and $\"y=-0.122.\"$ .

Solution :

Critical values are : $\"x=0.71\"$ and $\"y=-0.122.\"$ .