Question

The function,$\mathrm{f}\left(\mathrm{x}\right)=(3\mathrm{x}-7)·{\mathrm{x}}^{\frac{2}{3}}$, $\mathrm{x}\in \mathrm{R}$ is increasing for all $\mathrm{x}$ lying in:

• A. $\left(-\infty ,-\frac{14}{15}\right)\cup \left(0,\infty \right)$
• B. $\left(-\infty ,\frac{14}{15}\right)$
• C. $\left(-\infty ,0\right)\cup \left(\frac{14}{15},\infty \right)$
• D. $\left(-\infty ,0\right)\cup \left(\frac{3}{7},\infty \right)$

Answer 1

The correct option is C

$\left(-\infty ,0\right)\cup \left(\frac{14}{15},\infty \right)$

Explanation for correct option

Step 1. Determine the interval of $\mathrm{x}$.

The given function is, $\mathrm{f}\left(\mathrm{x}\right)=(3\mathrm{x}-7)·{\mathrm{x}}^{\frac{2}{3}}$

As we know that, if $\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right)$ is increasing then function $\mathrm{f}\text{'}\left(\mathrm{x}\right)$ should be greater than $0$.

We can rewrite the function as,

$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{\frac{2}{3}}·(3\mathrm{x}-7)$

Apply the distributive law $\mathit{a}\mathbf{(}\mathbf{b}\mathbf{-}\mathbf{c}\mathbf{)}\mathbf{=}\mathbf{ab}\mathbf{-}\mathbf{ac}$, where $\mathrm{a},\mathrm{b}$ and $\mathrm{c}$ are real numbers.

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)={\mathrm{x}}^{\frac{2}{3}}\times 3\mathrm{x}-{\mathrm{x}}^{\frac{2}{3}}\times 7$

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=3{\mathrm{x}}^{\frac{5}{3}}-7{\mathrm{x}}^{\frac{2}{3}}$

Step 2 Use Sum/Difference rule.

We will simplify further equation by using sum/Difference rule $\left(f\pm g\right)\mathbf{\text{'}}\mathbf{=}\mathbf{f}\mathbf{\text{'}}\mathbf{\pm }\mathbf{g}\mathbf{\text{'}}$, where $\mathrm{f}$ and $\mathrm{g}$ are differentiable function.

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(3{\mathrm{x}}^{\frac{5}{3}}\right)-\frac{\mathrm{d}}{\mathrm{dx}}\left(7{\mathrm{x}}^{\frac{2}{3}}\right)$

Take the constant out $\left(a\times f\right)\mathbf{\text{'}}\mathbf{=}\mathbf{a}\mathbf{\times }\mathbf{f}\mathbf{\text{'}}$.

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=3\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{\frac{5}{3}}\right)-7\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{\frac{2}{3}}\right)$

Apply the power rule $\frac{\mathbf{d}}{\mathbf{dx}}\left(x^a\right)\mathbf{=}\mathbf{a}\mathbf{\times }{\mathbf{x}}^{\mathbf{a}\mathbf{-}\mathbf{1}}$.

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=3\times \frac{5}{3}{\mathrm{x}}^{\frac{5}{3}-1}-7\times \frac{2}{3}{\mathrm{x}}^{\frac{2}{3}-1}$

Use exponent rule ${\mathbf{a}}^{\mathbf{-}\mathbf{b}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{a}}^{\mathbf{b}}}$.

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=3·\frac{5}{3}{\mathrm{x}}^{\frac{2}{3}}-7·\frac{2}{3}·\frac{1}{{\mathrm{x}}^{{\displaystyle \frac{1}{3}}}}$

Use fraction rule $\mathbf{a}\mathbf{·}\frac{\mathbf{b}}{\mathbf{c}}\mathbf{·}\frac{\mathbf{d}}{\mathbf{e}}\mathbf{=}\frac{\mathbf{a}\mathbf{\times }\mathbf{b}\mathbf{\times }\mathbf{d}}{\mathbf{c}\mathbf{\times }\mathbf{e}}$.

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=5{\mathrm{x}}^{\frac{2}{3}}-\frac{7\times 2\times 1}{3{\mathrm{x}}^{{\displaystyle \frac{1}{3}}}}$

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=5{\mathrm{x}}^{\frac{2}{3}}-\frac{14}{3{\mathrm{x}}^{{\displaystyle \frac{1}{3}}}}$

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{15\mathrm{x}-14}{3{\mathrm{x}}^{{\displaystyle \frac{1}{3}}}}>0$

Therefore, $\mathrm{f}\text{'}\left(\mathrm{x}\right)>0\forall \mathrm{x}\in \left(-\infty ,0\right)\cup \left(\frac{14}{15},\infty \right)$

Hence, Option $\left(\mathrm{C}\right)$ is correct answer.