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The rate of change of function $f\left( x \right)=3{{x}^{5}}-5{{x}^{3}}+5x-7$ is minimum when:\begin{align} & A.x=0 \\ & B.x=\dfrac{1}{\sqrt{2}} \\ & C.x=-\dfrac{1}{\sqrt{2}} \\ & D.x=\pm \dfrac{1}{\sqrt{2}} \\ \end{align}

Last updated date: 17th Jul 2024
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Hint: In order to find the rate of change of function $f\left( x \right)=3{{x}^{5}}-5{{x}^{3}}+5x-7$, firstly, we must consider a variable as rate. Then we must perform differentiation upon the given function. We obtain an expression. Since we are asked the minimum rate, we will be performing differentiation upon the obtained expression. Upon equating the first obtained expression to zero and solving it gives us the required rate.

Complete step-by-step solution:
Now let us learn about differentiation. We find the derivative of a function using differentiation. It is nothing but defining the rate of change of function. The sum rule of the differentiation says that the sum of the function is the sum of derivatives. While differentiating, two variables are considered. Generally, $x$ and $y$ are considered and is denoted as $\dfrac{dy}{dx}$. There exist a number of rules to find the derivatives of many functions. The inverse process of differentiation is integration.
Now let us find the rate of change of function $f\left( x \right)=3{{x}^{5}}-5{{x}^{3}}+5x-7$when it is minimum.
Firstly, let us consider a variable denoting the rate of change of function i.e. $v$
$\Rightarrow v=\dfrac{dv}{dx}$
Now let us find the derivative of the given function.
$\Rightarrow v=f’\left( x \right)=15{{x}^{4}}-15{{x}^{2}}+5$
Upon differentiating again, we get
$\dfrac{dv}{dx}=60{{x}^{3}}-30x$
$\Rightarrow \dfrac{{{d}^{2}}v}{{{d}^{2}}x}=180{{x}^{2}}-30$
Now let us consider, $\dfrac{dv}{dx}=0$
We have our $\dfrac{dv}{dx}$ as $60{{x}^{3}}-30x$.
So upon substituting, we get
$60{{x}^{3}}-30x=0$
Upon solving this,
We obtain $x=\pm \dfrac{1}{\sqrt{2}},0$
Now let us check, at which value the rate is minimum.
At $x=0$,
We have $\dfrac{{{d}^{2}}v}{{{d}^{2}}x}=180{{\left( 0 \right)}^{2}}-30=-30<0$
At $x=\pm \dfrac{1}{\sqrt{2}}$,
We have $\dfrac{{{d}^{2}}v}{{{d}^{2}}x}=180{{\left( \pm \dfrac{1}{\sqrt{2}} \right)}^{2}}-30=60>0$
$\therefore$$f’\left( x \right)$ i.e. $v$ is minimum at $x=\pm \dfrac{1}{\sqrt{2}}$.
Hence option D is correct.

Note: In order to find the rate of a function, we must always have a note that the rate of a function is found for the first derivative when unmentioned. And the obtained values will be substituted in the second derivatives if we are supposed to check the minimum or the maximum value.