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# 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 $R$

Last updated date: 08th Sep 2024
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Hint: 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.

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
\begin{align} & \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{align}
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{align} & D={{b}^{2}}-4ac \\ & ={{6}^{2}}-4\left( 3 \right)\left( 5 \right) \\ & =36-60 \\ & =-14 \end{align}
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.