Calculus Tutor: None Selected Time limit: 1 Day

Jun 30th, 2015

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.

Jun 30th, 2015

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Jun 30th, 2015
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Jun 30th, 2015
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