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Calculus

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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)^1/21]
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)^1/21 + 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.
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