Question

The minimum value of $f\left( x \right)={{e}^{\left( {{x}^{4}}-{{x}^{3}}+{{x}^{2}} \right)}}$ is

Answer 1

For these kinds of functions, we use differentiation and it’s first rule to find the minimum value. We know that ${{e}^{x}}$ is an increasing function. So at the maximum value of $x$, it will have its maximum value too. And at the minimum value of $x$, it will have its minimum value too. It will peak and lower with respect to its power since it is an increasing function. So now we just have to differentiate the power of $e$ here and find out the value of $x$ at which it is minimum.

Complete step by step solution:
So we should use both the rules of differentiation.
Let us assume that $h\left( x \right)={{x}^{4}}-{{x}^{3}}+{{x}^{2}}$ .
We know the formula of differentiation of ${{x}^{n}}$. It is $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$ .
Let us use this formula and differentiate it.
Upon doing so, we get the following :
$\begin{aligned} & \Rightarrow h\left( x \right)={{x}^{4}}-{{x}^{3}}+{{x}^{2}} \\\ & \Rightarrow h’\left( x \right)=4{{x}^{3}}-3{{x}^{2}}+2x=0 \\\ \end{aligned}$
The first rule of differentiation is that we can equate the first derivative of a function to $0$.
We already did that in the previous step. Now let us take $x$ common out of all the terms to find at what will $h\left( x \right)$ have peak values, either minimum or maximum.
Upon doing so, we get the following :
$\begin{aligned} & \Rightarrow h\left( x \right)={{x}^{4}}-{{x}^{3}}+{{x}^{2}} \\\ & \Rightarrow h'\left( x \right)=4{{x}^{3}}-3{{x}^{2}}+2x=0 \\\ & \Rightarrow h'\left( x \right)=x\left( 4{{x}^{2}}-3x+2 \right)=0 \\\ \end{aligned}$
So either$~~x=0$ or $4{{x}^{2}}-3x+2=0$
Let us first see if we can get any values of $x$ from $4{{x}^{2}}-3x+2$. So we will check the discriminant first.
Discriminant (D) = $\sqrt{{{b}^{2}}-4ac}$ where $a$ is the coefficient of ${{x}^{2}}$, $b$ is the coefficient of $x$ and $c$ is the constant.
Upon substituting , we get the following :
$\Rightarrow$ Discriminant (D) = $\sqrt{{{b}^{2}}-4ac}=\sqrt{{{\left( 3 \right)}^{2}}-4\times 4\times 2}=\sqrt{9-32}=\sqrt{-23}<0$
Since it’s Discriminant (D) is less than zero, it will not give us any real values of $x$. So the only other value of $x$, that we are remaining with is$~~x=0$.
Now we have to check whether we will get a maximum value of the minimum value of the function $f\left( x \right)$ when we substitute$~~x=0$.
For this we need the second rule of differentiation.
Let us differentiate it again.
Upon doing so, we get the following :
$\begin{aligned} & \Rightarrow h\left( x \right)={{x}^{4}}-{{x}^{3}}+{{x}^{2}} \\\ & \Rightarrow h'\left( x \right)=4{{x}^{3}}-3{{x}^{2}}+2x=0 \\\ & \Rightarrow h''\left( x \right)=12{{x}^{2}}-6x+2 \\\ \end{aligned}$
Let us substitute $x=0$.
Upon doing so , we get the following :
$\begin{aligned} & \Rightarrow h\left( x \right)={{x}^{4}}-{{x}^{3}}+{{x}^{2}} \\\ & \Rightarrow h'\left( x \right)=4{{x}^{3}}-3{{x}^{2}}+2x=0 \\\ & \Rightarrow h''\left( x \right)=12{{x}^{2}}-6x+2 \\\ & \Rightarrow h''\left( 0 \right)=2>0 \\\ \end{aligned}$
Since $h''\left( 0 \right)>0$ , we will definitely get the minimum value of function $h\left( x \right)$ at $x=0$ .
The minimum value of the function $h\left( x \right)={{x}^{4}}-{{x}^{3}}+{{x}^{2}}$is attained when $x=0$ . And the minimum value of $h\left( x \right)={{x}^{4}}-{{x}^{3}}+{{x}^{2}}$ is $h\left( 0 \right)={{\left( 0 \right)}^{4}}-{{\left( 0 \right)}^{3}}+{{\left( 0 \right)}^{2}}=0$ .
The minimum value of the function $f\left( x \right)={{e}^{\left( {{x}^{4}}-{{x}^{3}}+{{x}^{2}} \right)}}$ would be ${{e}^{0}}=1$ .

Hence, The minimum value of $f\left( x \right)={{e}^{\left( {{x}^{4}}-{{x}^{3}}+{{x}^{2}} \right)}}$ is $1$ .

Note: It is very important to remember the rules of differentiation. We should also remember the different ways of differentiation so as to complete a question quickly. We should also remember the derivatives of different functions. We should be careful while solving as there is a lot of scope for calculation errors. Calculus is a vast subject and a lot of practice is required. We should do various problems to get a hold and feel of the subject.