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

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.

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
\[\begin{align}
  & \dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\left( {{x}^{3}}-3{{x}^{2}}+5x+7 \right) \\
 & =3{{x}^{2}}-6x+5+0 \\
 & =3{{x}^{2}}-6x+5
\end{align}\]
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.