Question

The function f is defined by $f\left( x \right)={{x}^{3}}-3{{x}^{2}}+5x+7$. Find the nature of the function.
(a) decreasing in $R$
(b) decreasing in $\left( 0,\infty \right)$ and increasing in $\left( -\infty ,0 \right)$
(c) increasing in $\left( 0,\infty \right)$ and decreasing in $\left( -\infty ,0 \right)$
(d) increasing in $R$

Answer 1

In this question, we have a function $f$ is defined by $f\left( x \right)={{x}^{3}}-3{{x}^{2}}+5x+7$. We will first find the derivative $\dfrac{d}{dx}f\left( x \right)$ of the given function. We will then check if $\dfrac{d}{dx}f\left( x \right)$ greater than zero or it is less than zero. Now using the first derivative test for a function $f$ which states that if $\dfrac{d}{dx}f\left( x \right)$ greater than zero, then the function is decreasing and if $\dfrac{d}{dx}f\left( x \right)$ less than zero, then the function is increasing. We will then get our desired answer.

Complete step by step answer:
We are given a function $f$ is defined by $f\left( x \right)={{x}^{3}}-3{{x}^{2}}+5x+7$.
We will now find the derivative $\dfrac{d}{dx}f\left( x \right)$ of the given function $f\left( x \right)={{x}^{3}}-3{{x}^{2}}+5x+7$.
The derivative of the function $f\left( x \right)={{x}^{3}}-3{{x}^{2}}+5x+7$ is given by

& \dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\left( {{x}^{3}}-3{{x}^{2}}+5x+7 \right) \\\ & =3{{x}^{2}}-6x+5+0 \\\ & =3{{x}^{2}}-6x+5 \end{aligned}$$ Therefore we have $$\dfrac{d}{dx}f\left( x \right)=3{{x}^{2}}-6x+5$$. Now we can see that the derivative $$\dfrac{d}{dx}f\left( x \right)$$ of the given function$$f\left( x \right)={{x}^{3}}-3{{x}^{2}}+5x+7$$ is a quadratic polynomial. That is a polynomial of degree 2. Now for any quadratic polynomial $$a{{x}^{2}}-bx+c$$, the discriminant is given by $$D={{b}^{2}}-4ac$$ On comparing the derivative $$\dfrac{d}{dx}f\left( x \right)=3{{x}^{2}}-6x+5$$ with the quadratic polynomial $$a{{x}^{2}}-bx+c$$,we will have $$a=3,b=-6$$ and $$c=5$$ Therefore the discriminant of the quadratic equations $$\dfrac{d}{dx}f\left( x \right)=3{{x}^{2}}-6x+5$$ is given by $$\begin{aligned} & D={{b}^{2}}-4ac \\\ & ={{6}^{2}}-4\left( 3 \right)\left( 5 \right) \\\ & =36-60 \\\ & =-14 \end{aligned}$$ Therefore $$D<0$$ Since we know that for a quadratic polynomial $$a{{x}^{2}}-bx+c$$, if the discriminant $$D={{b}^{2}}-4ac$$ is greater than zero then the polynomial $$a{{x}^{2}}-bx+c$$ is less than zero for all positive real values of $$x$$ and if $$D={{b}^{2}}-4ac$$ is less than zero then the polynomial $$a{{x}^{2}}-bx+c$$ is greater than zero for all positive real values of $$x$$. Now here for the $$\dfrac{d}{dx}f\left( x \right)=3{{x}^{2}}-6x+5$$, the discriminant is less than zero. Therefore we have that $$\dfrac{d}{dx}f\left( x \right)=3{{x}^{2}}-6x+5$$ is greater than zero for all $$x\in {{R}^{+}}$$. Also since we know that $$\dfrac{d}{dx}f\left( x \right)=3{{x}^{2}}-6x+5$$ is greater than zero for all $$x\in {{R}^{-}}$$. Therefore we have that the derivative $$\dfrac{d}{dx}f\left( x \right)=3{{x}^{2}}-6x+5$$ is greater than zero for all $$x\in R$$. Now using the first derivative test for a function $$f$$ which states that if $$\dfrac{d}{dx}f\left( x \right)$$ greater than zero, then the function is decreasing and if $$\dfrac{d}{dx}f\left( x \right)$$ less than zero, then the function is increasing. We get that the function given by $$f\left( x \right)={{x}^{3}}-3{{x}^{2}}+5x+7$$ is increasing on $$R$$. **So, the correct answer is “Option D”.** **Note:** In this problem, in order to determine the nature of the given function $$f\left( x \right)={{x}^{3}}-3{{x}^{2}}+5x+7$$ we have to actually find the function increasing or decreasing on the real line. And this can be done directly using the first derivative test for a function $$f$$ which states that if $$\dfrac{d}{dx}f\left( x \right)$$ greater than zero, then the function is decreasing and if $$\dfrac{d}{dx}f\left( x \right)$$ less than zero, then the function is increasing.