Answer:
[tex]\textsf{Increasing function}: \quad \left[0, \dfrac{1}{16}\right)[/tex]
[tex]\textsf{Decreasing function}: \quad \left(\dfrac{1}{16}, \infty\right)[/tex]
Step-by-step explanation:
Given function:
[tex]f(x)=\sqrt{x}-2x[/tex]
As a negative number cannot be square rooted, the domain of the function is restricted to [0, ∞).
[tex]\textsf{A function is \textbf{increasing} when the \underline{gradient is positive}}\implies f'(x) > 0[/tex]
[tex]\textsf{A function is \textbf{decreasing} when the \underline{gradient is negative}} \implies f'(x) < 0[/tex]
Differentiating produces an algebraic expression for the gradient as a function of x. Therefore, differentiate the given function:
[tex]\begin{aligned}f(x) & = \sqrt{x}-2x\\& = x^{\frac{1}{2}}-2x\\\implies f'(x) & = \dfrac{1}{2}x^{\left(\frac{1}{2}-1\right)}-2x^{(1-1)}\\ & = \dfrac{1}{2}x^{-\frac{1}{2}}-2x^0\\ & = \dfrac{1}{2\sqrt{x}}-2\end{aligned}[/tex]
Increasing function
To find the interval where f(x) is increasing, set the differentiated function to more than zero and solve for x:
[tex]\begin{aligned}f'(x) & > 0\\\implies \dfrac{1}{2\sqrt{x}}-2 & > 0\\\dfrac{1}{2\sqrt{x}} & > 2\\\ \dfrac{1}{2} & > 2\sqrt{x}\\\dfrac{1}{2 \cdot 2} & > \sqrt{x}\\ \dfrac{1}{4} & > \sqrt{x}\\ \sqrt{x} & < \dfrac{1}{4}\\\left(\sqrt{x}\right)^2 & < \left(\dfrac{1}{4}\right)^2\\ x & < \dfrac{1}{16}\end{aligned}[/tex]
As the domain is restricted, the function is increasing when:
[tex]\textsf{Solution}: \quad 0 \leq x < \dfrac{1}{16}[/tex]
[tex]\textsf{Interval notation}: \quad \left[0, \dfrac{1}{16}\right)[/tex]
Decreasing function
To find the interval where f(x) is decreasing, set the differentiated function to less than zero and solve for x:
[tex]\begin{aligned}f'(x) & < 0\\\implies \dfrac{1}{2\sqrt{x}}-2 & < 0\\\dfrac{1}{2\sqrt{x}} & < 2\\\ \dfrac{1}{2} & < 2\sqrt{x}\\\dfrac{1}{2 \cdot 2} & < \sqrt{x}\\ \dfrac{1}{4} & < \sqrt{x}\\ \sqrt{x} & > \dfrac{1}{4}\\\left(\sqrt{x}\right)^2 & > \left(\dfrac{1}{4}\right)^2\\ x & > \dfrac{1}{16}\end{aligned}[/tex]
Therefore, the function is decreasing when:
[tex]\textsf{Solution}: \quad x > \dfrac{1}{16}[/tex]
[tex]\textsf{Interval notation}: \quad \left(\dfrac{1}{16}, \infty\right)[/tex]
Note: The answer quoted in the original question is incorrect for the quoted function (please refer to the attached graph for proof).
Differentiation Rules
[tex]\boxed{\begin{minipage}{4.8 cm}\underline{Differentiating $ax^n$}\\\\If $y=ax^n$, then $\dfrac{\text{d}y}{\text{d}x}=nax^{n-1}$\\\end{minipage}}[/tex]
