Question

Where does f(x)=x+√1-x, 0 < x < 1 increase or decrease? Find its maximum and minimum values.

Answer 1

function, f(x) = x + √(1 - x), where 0 < x < 1

differentiating f(x) with respect to x,

f'(x) = 1 + 1/2√(1 - x) × (-1)

= 1 - 1/2√(1 - x)

function is increasing when f'(x) > 0

i.e., 1 - 1/2√(1 - x) > 0

or, 1 > 1/2√(1 - x)

squaring both sides,

or, 1 > 1/4(1 - x)

or, 4 > 1/(1 - x)

or, 1/(1 - x) - 4 < 0

or, (1 - 4 + 4x)/(1 - x) < 0

or, (4x - 3)/(1 - x) < 0

or, (4x - 3)/(x - 1) > 0

or, x > 1 or, x < 3/4

but domain is 0 < x < 1

so, 0 < x < 3/4

hence, function is increasing on (0, 3/4).

similarly, function is decreasing when f'(x) < 0

i.e., 1 - 1/2√(1 - x) < 0

we get, 3/4 < x < 1

in interval, (3/4, 1) function is decreasing.

now, f'(x) = 1 - 1/2√(1 - x) = 0

or, 1 = 1/2√(1 - x)

or, 4(1 - x) = 1

or, 4 - 4x = 1

or, x = 3/4

f"(x) = 0 - 1/4(1 - x)^{3/2}

at x = 3/4, f"(3/4) < 0

hence, maximum value of f(x) at x = 3/4

and maximum value of f(x) = f(3/4) = 3/4 + √(1 - 3/4) = 3/4 + 1/2 = 5/4

[note : minimum value of this function doesn't exist. ]