Check the solution for me!Calculus 1. Extrema!

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.

And the square root of x is being multiplied by (x-3). the "x" and "x-3" are not all inside the square root, jut the one "x"

So someone told me this:
′(x)=x√.1+[(x−3)/2√(x)] =[2x+x−3/2√(x)]= [3x−3/2√(x)]

f'(x)=0 gives x=1
f′′(x)=1/2 ∗[√(x)∗3−(3x−3)1/2(²√x)] / x

at x=1
f′′(x)=(1/2)∗√(1)∗3−0 / 2∗√(1)= (3/4)>0

Hence f(x) is minimum at x=1
for increasing f'(x)>0
for decreasing f'(x)<0

This seems sort of confusing to me! Is this right? If so, what else would I do after?? This threw me off! Please help!

The domain of this function only exists when x is greater than or equal to 0 (domain of sqrt(x) restricts us)

f(x) = (x - 3) * sqrt(x)
f'(x) = (x - 3) * (1/2) * x^(-1/2) + sqrt(x)
f'(x) = 0
0 = (x - 3) / (2 * sqrt(x)) + sqrt(x)
(3 - x) / (2 * sqrt(x)) = sqrt(x)
3 - x = 2 * sqrt(x)
0 = x + 2 * sqrt(x) - 3
sqrt(x) = (-2 +/- sqrt(4 + 12)) / 2
sqrt(x) = (-2 +/- 4) / 2
sqrt(x) = -6/2 , 2/2
sqrt(x) = -3 , 1
Remember, our domain is restricted
sqrt(x) = 1
x = 1

So your domains are (0 , 1) and (1 , inf)

f'(x) = (x - 3) / (2 * sqrt(x)) + sqrt(x)
f'(1/4) = (1/4 - 3) / (2 * sqrt(1/4)) + sqrt(1/4)
f'(1/4) = (-11/4) / (2 * (1/2)) + 1/2
f'(1/4) = -11/4 + 1/2
f'(1/4) = -11/4 + 2/4
f'(1/4) = -9/4

f'(4) = (4 - 3) / (2 * sqrt(4)) + sqrt(4)
f'(4) = 1 / (2 * 2) + 2
f'(4) = 1/4 + 2
f'(4) = 9/4

So, from 0 to 1, f(x) is decreasing and from 1 onward f(x) is increasing. That makes f(1) a minimum point

f(1) = (1 - 3) * sqrt(1) = -2 * 1 = -2

(1 , -2) is your only extrema

Yes, you are correct.

so 3x-3 < 0 for f'(x) to be decreasing and x < 3

for f'(x) to be increasing, x > 3