Asked by Abdulrahim m umar
find the maximum and minimum value of function using first and second driver ties test f(x)=x+√(1-x).
Answers
Answered by
Reiny
your school subject is : math
not: Abubakar tafawa balewa university bauchi ATBU
y = x + (1-x)^(1/2)
y' = 1 + (1/2)(1-x)^(-1/2)(-1)
= 1 - (1/2)(1-x)^(-1/2)
= 0 for a max/min
1 = (1/2) (1/√(1-x)
√(1-x) = 1/2
1-x = 1/4
4-4x=1
x = 3/4
f(3/4) = 3/4 + √(1-3/4)
= 3/4 + 1/2 = 5/4 ----> max
y'' = 0 + (1/4)(1-x)^(-3/2)(-1)
= -(1/4)(1/(1-x)^(3/2) ) )
at x = 3/4
y'' = -(1/4)(1/√1/4)) = negative
so 5/4 is a maximum when x = 3/4
not: Abubakar tafawa balewa university bauchi ATBU
y = x + (1-x)^(1/2)
y' = 1 + (1/2)(1-x)^(-1/2)(-1)
= 1 - (1/2)(1-x)^(-1/2)
= 0 for a max/min
1 = (1/2) (1/√(1-x)
√(1-x) = 1/2
1-x = 1/4
4-4x=1
x = 3/4
f(3/4) = 3/4 + √(1-3/4)
= 3/4 + 1/2 = 5/4 ----> max
y'' = 0 + (1/4)(1-x)^(-3/2)(-1)
= -(1/4)(1/(1-x)^(3/2) ) )
at x = 3/4
y'' = -(1/4)(1/√1/4)) = negative
so 5/4 is a maximum when x = 3/4
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