Question

For the function f(x)=√( x)(x-3) , find the open intervals on which the function is
increasing or decreasing, and find the relative extrema.

I have tried all that my notes can provide and OS is my last resort. If someone could show me a step by step solution on how to get this done I'd appreciate it so muchh! I really do not know how to continue on with this or if I am in the right direction!

Answer 1

is the function
\[f(x)=\sqrt{x}(x-3)\]?

Answer 2

or is it all under the radical like
\[f(x)=\sqrt{x(x-3)}\]?

Answer 3

the first one :)

Answer 4

Do I start with finding the 2nd derivative? I just feel clueless at this point.

Answer 5

\[f'(x)=\sqrt{x}.1+\frac{ x-3 }{2\sqrt{x} }=\frac{ 2x+x-3 }{2\sqrt{x} }=\frac{ 3x-3 }{ 2 \sqrt{x} }\]
f'(x)=0 gives x=1
\[f \prime \prime \left( x \right)=\frac{ 1 }{2 } *\frac{\sqrt{x}*3-(3x-3)\frac{ 1 }{2 \sqrt{x}} }{x }\]
at x=1
\[f \prime \prime(x)=\frac{ 1 }{2}*\frac{ \sqrt{1}*3-0 }{ 2*\sqrt{1} }=\frac{ 3 }{4 }>0\]
Hence f(x) is minimum at x=1
for increasing f'(x)>0
for decreasing f'(x)<0
solve.

Answer 6

remember here \[x \ge 0\]