Find the correct option.
The function \[f(x) = (3x - 7){x^{\dfrac{2}{3}}},x \in R\] is increasing for all \[x\] lying in
A. \[\left( { - \infty , - \dfrac{{14}}{{15}}} \right) \cup (0,\infty )\]
B. \[\left( { - \infty , - \dfrac{{14}}{{15}}} \right)\]
C. \[( - \infty ,0) \cup \left( { - \infty , - \dfrac{{14}}{{15}}} \right)\]
D. \[( - \infty ,0) \cup \left( { - \infty , - \dfrac{{14}}{{15}}} \right)\]

Hint: The first derivative of the function \[f(x)\] will be found and inequality will be established using the rule for an increasing function with that calculated derivation and zero to solve for \[x\] and choose the correct option.

Formulae Used: The following formulas will be used to solve this problem:
1. \[\dfrac{{d(uv)}}{{dx}} = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\], where, \[u\] and \[v\] are functions of \[x\].
2. \[\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}},n \in R\].
3. The function \[f(x)\] will be an increasing function, if \[f'(x)\] is greater than zero.

Complete step-by-step solution:
We have been given that \[f(x) = (3x - 7){x^{\dfrac{2}{3}}},x \in R\] .
We will first find the first derivative of the function.
Let, \[y = (3x - 7){x^{\dfrac{2}{3}}}\]
\[u = (3x - 7)\] and \[v = {x^{\dfrac{2}{3}}}\]
Then, \[\dfrac{{dy}}{{dx}} = \dfrac{{d(uv)}}{{dx}}\]
\[ = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\] [Applying the formula]
\[ = (3x - 7)\dfrac{{d({x^{\dfrac{2}{3}}})}}{{dx}} + {x^{\dfrac{2}{3}}}\dfrac{{d(3x - 7)}}{{dx}}\] [Substituting the value of \[u\] and \[v\]]
\[ = (3x - 7) \times \dfrac{2}{3} \times {x^{\left( {\dfrac{2}{3} - 1} \right)}} + {x^{\dfrac{2}{3}}} \times 3\] [Using the formula \[\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}},n \in R\] and considering \[n = \dfrac{2}{3}\] ]
Further solving the above,
\[f'(x) = \dfrac{{2(3x - 7)}}{{3{x^{\dfrac{1}{3}}}}} + 3{x^{\dfrac{2}{3}}}\]
\[ = \dfrac{{6x - 14 + 9x}}{{3{x^{\dfrac{1}{3}}}}}\]
\[ = \dfrac{{15x - 14}}{{3{x^{\dfrac{1}{3}}}}}\]
If \[f(x)\] is an increasing function, then \[f'(x)\] is greater than zero.
So, \[\dfrac{{15x - 14}}{{3{x^{\dfrac{1}{3}}}}} > 0\]
Let, \[\dfrac{{15x - 14}}{{3{x^{\dfrac{1}{3}}}}} = \dfrac{a}{b}\] , where \[(15x - 14) = a,3{x^{\dfrac{2}{3}}} = b\] and \[a,b \in R\] .
Case 1 if \[\dfrac{a}{b} > 0\] , then \[a > 0\] and \[b > 0\] .
Case 2 if \[\dfrac{a}{b} > 0\] , then \[a < 0\] and \[b < 0\] .
We will examine the above two cases one by one.
From case 1, \[(15x - 14) > 0\] and \[3{x^{\dfrac{2}{3}}} > 0\]
\[ \Rightarrow 15x > 14\] and \[{x^{\dfrac{2}{3}}} > 0\]
\[ \Rightarrow x > \dfrac{{14}}{{15}}\] and \[x > 0\]
\[ \Rightarrow x \in \left( {\dfrac{{14}}{{15}},\infty } \right)\]
From case 2, \[(15x - 14) < 0\] and \[3{x^{\dfrac{2}{3}}} < 0\]
\[ \Rightarrow (15x - 14) < 0\] and \[3{x^{\dfrac{2}{3}}} < 0\]
\[ \Rightarrow 15x < 14\] and \[{x^{\dfrac{2}{3}}} < 0\]
\[ \Rightarrow x < \dfrac{{14}}{{15}}\] and \[x < 0\]
\[ \Rightarrow x \in ( - \infty ,0)\]
From the above, it is clear that \[x \in \left( { - \infty ,0} \right) \cup \left( {\dfrac{{14}}{{15}},\infty } \right)\] .

Hence, option C. is the correct answer.

Note: The addition and subtraction of the same number or expression to each side of an inequation does not change the inequality. Multiplying and dividing each side of an inequation by the same positive number also does not change the inequality, but, multiplying and dividing each side of an inequation by the same negative number reverses the inequality.