The function $f\left( x \right) = 2{x^3} - 3{x^2} + 90x + 174$ is increasing in the interval
1. $\dfrac{1}{2} < x < 1$
2. $\dfrac{1}{2} < x < 2$
3. $3 < x < \dfrac{{59}}{4}$
4. $ - \infty < x < \infty $

Hint: In this problem, we have to find the increasing interval of given function $f\left( x \right) = 2{x^3} - 3{x^2} + 90x + 174$. First step is to find the derivative of given function $f\left( x \right)$ with respect to $x$. Then, check for which values the first derivative is greater than zero because that interval will be the increasing interval of any function.

Formula Used:
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$

Complete step by step Solution:
Given that,
$f\left( x \right) = 2{x^3} - 3{x^2} + 90x + 174$
Differentiate $f\left( x \right)$ with respect to $x$,
$f'\left( x \right) = 6{x^2} - 6x + 90$
$6{x^2} - 6x + 90 > 0$
${x^2} - x + 15 > 0$
Here, $f'\left( x \right) > 0\forall x$
$\therefore f\left( x \right)$ is increasing for $ - \infty < x < \infty $

Hence, the correct option is (4).

Note:In such a question, if we are given any function and we have to check in which interval the function is increasing or decreasing always find the first derivative of that function. Then, for an increasing interval should satisfy $f'\left( x \right) > 0$ this condition. Similarly for decreasing interval it will be $f'\left( x \right) < 0$.