Where does f(x) = x + sqrt{1 - x}, 0

Question

(A) To find where the function $f(x) = x + \sqrt{1 - x}$ decreases,we find its derivative $f'(x)$.$f'(x) = \frac{d}{dx}(x) + \frac{d}{dx}((1 - x)^{1/2

Vedclass · Accepted Answer

$\left(\frac{3}{4}, 1\right)$

Vedclass · Answer

(A) To find where the function $f(x) = x + \sqrt{1 - x}$ decreases,we find its derivative $f'(x)$.$f'(x) = \frac{d}{dx}(x) + \frac{d}{dx}((1 - x)^{1/2}) = 1 + \frac{1}{2}(1 - x)^{-1/2}(-1) = 1 - \frac{1}{2\sqrt{1 - x}}$.For the function to decrease,we set $f'(x) $1 - \frac{1}{2\sqrt{1 - x}} Since $0 $2\sqrt{1 - x} Squaring both sides:$1 - x  \frac{3}{4}$.Given the domain $0 Thus,the correct option is $A$.

Where does $f(x) = x + \sqrt{1 - x}, 0 < x < 1$ decrease?

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