\ \

\

Step 1 :

\

The function is \"\"

\

Domain of the function : \ \

\

Since there should not be any negative numbers in the square root, \"\" \ \

\

\"\" \ \

\

The domain is \"\".

\

Step 2 :

\

Let \"\"

\

Apply derivative on each side with respect to x.

\

\"\"

\

Apply the product rule of derivative:\"\"

\

\"\"

\

\"\"

\

Step 3 :

\

Determination of critical points:

\

Since \"\" is a root function, it is continuous on its domain \"\".

\

The critical points exists when \"\".

\

Equate \"\" to zero:

\

\"\"

\

The critical points are \"\"

\

Consider the test intervals as \"\" and \"\". \ \

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
Interval Test Value Sign of \"\"Conclusion
\"\" \"\" \

\"\"

\
Decreasing
\"\" \"\" \

\"\"

\
Increasing
\"\" \"\" \

\"\"

\
Decreasing
\

Thus, The function is increasing on the interval \"\". \ \

\

And The function is decreasing on the intervals \"\" and \"\".

\

Solution :

\

The function \"\" is increasing on the interval \"\". \ \

\

The function \"\" is decreasing on the intervals \"\" and \"\".