Question

The function $f\left(x\right)=2x^3-3x^2+90x+174$ is increasing in the interval.

• A. $\frac{1}{2}<x<1$
• B. $\frac{1}{2}<x<2$
• C. $3<x<\frac{59}{4}$
• D. $-\infty <x<\infty$

Answer 1

The correct option is D

$-\infty <x<\infty$

Explanation for the correct option

Given equation is $f\left(x\right)=2x^3-3x^2+90x+174$

An equation is increasing at $x$ if $f\text{'}\left(x\right)>0$.

We have,
$\begin{array}{c}f\text{'}\left(x\right)=\frac{d}{dx}\left(2x^3-3x^2+90x+174\right)\\ =6x^2-6x+90\\ =6\left(x^2-x+15\right)\end{array}$

$\therefore f\text{'}\left(x\right)$ is increasing if $x^2-x+15>0$

$x^2-x+15>0$ is always true because $x^2>x$. Also because, a positive greater number subtracted by a smaller number plus a positive number is always positive.
So $x^2-x+15>0$

Thus, $f\text{'}\left(x\right)>0\forall x$
$\Rightarrow f\text{'}\left(x\right)>0,-\infty <x<\infty$.

Hence, option(D) i.e. $-\infty <x<\infty$ is correct.