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?

Similar Questions

The length of the longest interval,in which the function $f(x) = 3\sin x - 4\sin^3 x$ is increasing,is

The equation $x^3+x-1=0$ has

If the function $f(x) = \int \frac{x + 3}{x^2 - 9x + 20} dx$, then $f(x)$ is

$f(x) = x^3 - 27x + 5$ is an increasing function,when

$A$ function $y = f(x)$ is given by $x = \frac{1}{1 + t^2}$ and $y = \frac{1}{t(1 + t^2)}$ for all $t > 0$. Then $f$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module