Please show all the work and make sure answer is correct

label Calculus
account_circle Unassigned
schedule 1 Day
account_balance_wallet $5

Jun 30th, 2015

Thank you for the opportunity to help you with your question!

1) Find the first derivative of f(x). So we have the following:

f(x) = x(x + 2)^1/2  --------> f '(x) = (x)'(x+2)^1/2 + x[ (x+2)^1/2 ]'  ----------> f '(x) = (1)(x+2)^1/2 + x[1/2(x+2)^1/2-1]

f '(x) = (x+2)^1/2 + x[1/2(x+2)^-1/2]  --------> f '(x) = (x+2)^1/2 + 1/2x(x+2)^-1/2

2) Then we equal it to zero and solve for x in order to find the critical numbers.

[ (x+2)^1/2 + 1/2x(x+2)^-1/2 = 0 ]*(x+2)^1/2 ----------> (x+2)^1/2(x+2)^1/2 + 1/2x(x+2)^-1/2(x+2)^1/2 = 0

(x+2)^1 + 1/2x(x+2)^0 = 0 --------->  x+2 + 1/2x = 0  ----------> 3/2x + 2 = 0  --------> 3/2x = - 2  -----> x = -2*2/3

x = - 4/3

3) After this, we write the ordered pair (x , y) by enterinf -4/3 in place of x into the original function like this:

f(-4/3) = (-4/3)(-4/3 + 2)^1/2 = (-4/3)(-4/3 + 6/3)^1/2 = (-4/3)(2/3)^1/2 = -1.089  ---------> (-4/3, -1.089)

4) Then we find the second derivative:

f ''(x) = 1/2(x+2)^1/2-1 + 1/2(x)'(x+2)^-1/2 + 1/2x*[(x+2)^-1/2]' = 1/2(x+2)^-1/2 + 1/2(x+2)^-1/2 -1/4x(x+2)^-3/2

f ''(x) = (x+2)^-1/2 - 1/4x(x+2)^-3/2

4) Finally, we enter the obtained critical number into the second derivative to determine whether or not it is a maximum or a minimum.

f ''(-4/3) = 1.43 > 0  then it's a local minimum. Remember:  if f ''(x) > 0 then it's a minimum and if f ''(x) < 0 then it is a maximum.

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 30th, 2015

Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.
Click to visit
The Notebank
...
Jun 30th, 2015
...
Jun 30th, 2015
Oct 19th, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer