Question

The function $f$ defined by $f\left(x\right)=4x^4-2x+1$ is increasing for

• A. $x<1$
• B. $x>0$
• C. $x<\frac{1}{2}$
• D. $x>\frac{1}{2}$

Answer 1

The correct option is D

$x>\frac{1}{2}$

Explanation for the correct option

Given function, $f\left(x\right)=4x^4-2x+1$.

Differentiate the given function with respect to $x$.

$\begin{array}{rcl}\frac{\mathrm{d}}{dx}\left[f\left(x\right)\right]& =& \frac{d}{dx}\left(4x^4-2x+1\right)\\ & =& \frac{d}{dx}\left(4x^4\right)+\frac{d}{dx}\left(-2x\right)+\frac{d}{dx}\left(1\right)\left[\because \frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{(}\mathbf{u}\mathbf{\pm }\mathbf{v}\mathbf{)}\mathbf{=}\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\mathbf{\pm }\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\right]\\ & =& \left(4\times 4x^3\right)-\left(2\times 1\right)+0\\ & =& 16x^3-2\end{array}$

Now, for increasing intervals $\mathit{f}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{>}\mathbf{0}$.

$$\begin{gathered} \Rightarrow 16x^3-2>0 \\ \Rightarrow x^3>\frac{2}{16} \\ \Rightarrow x^3>\frac{1}{8} \\ \Rightarrow x>\sqrt[3]{\frac{1}{8}} \\ \Rightarrow x>\frac{1}{2} \end{gathered}$$

Therefore, the function $f\left(x\right)=4x^4-2x+1$ is increasing for $x>\frac{1}{2}$.

Hence, $x>\frac{1}{2}$ and therefore the correct option is (D).