Question

Find the intervals where f(x) is increasing or decreasing: √x -2x. Ans: increasing (0, 1/4) Decreasing (1/4, ∞)

Answer 1

`\textsf{Increasing function}: \quad \left[0, (1)/(16)\right)`

`\textsf{Decreasing function}: \quad \left((1)/(16), \infty\right)`

Explanation:

Given function:

`f(x)=√(x)-2x`

As a negative number cannot be square rooted, the domain of the function is restricted to [0, ∞).

`\textsf{A function is \textbf{increasing} when the \underline{gradient is positive}}\implies f'(x) > 0`

`\textsf{A function is \textbf{decreasing} when the \underline{gradient is negative}} \implies f'(x) < 0`

Differentiating produces an algebraic expression for the gradient as a function of x. Therefore, differentiate the given function:

`\begin{aligned}f(x) & = √(x)-2x\\& = x^{(1)/(2)}-2x\\\implies f'(x) & = (1)/(2)x^{\left((1)/(2)-1\right)}-2x^((1-1))\\ & = (1)/(2)x^{-(1)/(2)}-2x^0\\ & = (1)/(2√(x))-2\end{aligned}`

Increasing function

To find the interval where f(x) is increasing, set the differentiated function to more than zero and solve for x:

`\begin{aligned}f'(x) & > 0\\\implies (1)/(2√(x))-2 & > 0\\(1)/(2√(x)) & > 2\\\ (1)/(2) & > 2√(x)\\(1)/(2 \cdot 2) & > √(x)\\ (1)/(4) & > √(x)\\ √(x) & < (1)/(4)\\\left(√(x)\right)^2 & < \left((1)/(4)\right)^2\\ x & < (1)/(16)\end{aligned}`

As the domain is restricted, the function is increasing when:

`\textsf{Solution}: \quad 0 \leq x < (1)/(16)`

`\textsf{Interval notation}: \quad \left[0, (1)/(16)\right)`

Decreasing function

To find the interval where f(x) is decreasing, set the differentiated function to less than zero and solve for x:

`\begin{aligned}f'(x) & < 0\\\implies (1)/(2√(x))-2 & < 0\\(1)/(2√(x)) & < 2\\\ (1)/(2) & < 2√(x)\\(1)/(2 \cdot 2) & < √(x)\\ (1)/(4) & < √(x)\\ √(x) & > (1)/(4)\\\left(√(x)\right)^2 & > \left((1)/(4)\right)^2\\ x & > (1)/(16)\end{aligned}`

Therefore, the function is decreasing when:

`\textsf{Solution}: \quad x > (1)/(16)`

`\textsf{Interval notation}: \quad \left((1)/(16), \infty\right)`

Note: The answer quoted in the original question is incorrect for the quoted function (please refer to the attached graph for proof).

Differentiation Rules

`\boxed{\begin{minipage}{4.8 cm}\underline{Differentiating $ax^n$}\\\\If $y=ax^n$, then $\frac{\text{d}y}{\text{d}x}=nax^(n-1)$\\\end{minipage}}`