Find the local maximum and minimum values of f using both the first and second derivative tests.?

Not sure how the second derivative helps, but if you get the first derivative and solve for its zeroes, it will give you the values of x that makes a min or max to the original function. Then solve for those values.

f(x) = x + √(1 - x)

f'(x) = 1 - 1 / [2√(1 - x)]

solve for zeroes:

0 = 1 - 1 / [2√(1 - x)]

-1 = -1 / [2√(1 - x)]

Multiply both sides by -√(1 - x)

√(1 - x) = 1/2

Square both sides:

1 - x = 1/4

Add x and subtract 1/4 from both sides:

x = 3/4

There is only one value that is a min or max.

f(x) = x + √(1 - x)

f(3/4) = 3/4 + √(1 - 3/4)

f(3/4) = 3/4 + √(1/4)

f(3/4) = 3/4 + 1/2

f(3/4) = 3/4 + 2/4

f(3/4) = 5/4

So the only local maximum is:

f(3/4) = 5/4

f'(x) = 1 - (1 / (2 * sqrt(1 - x)))

f''(x) = -1/(4(1 - x)^1.5))

To find extrema, set the derivative to zero:

1 - (1 / (2 * sqrt(1 - x))) = 0

(1 / (2 * sqrt(1 - x))) = 1

2 * sqrt(1 - x) = 1/1

2 * sqrt(1 - x) = 1

sqrt(1 - x) = 1/2

sqrt(1 - x) = 0.5

1 - x = 0.5^2

1 - x = 0.25

x = 1 - 0.25

x = 0.75

f(0.75) = 0.75 + sqrt(1 - 0.75)

f(0.75) = 0.75 + sqrt(0.25)

f(0.75) = 0.75 + 0.5

f(0.75) = 1.25