Question

The minimum value of $f\left(x\right)=2x^3-21x^2+36x-20$ is

• A. $-128$
• B. $-126$
• C. $-120$
• D. None of these

Answer 1

The correct option is A

$-128$

Find the minimum value of the given function

Given : $f\left(x\right)=2x^3-21x^2+36x-20$

Differentiate the function with respect to $x$,

$f\text{'}\left(x\right)=6x^2-42x+36$

For maximum or minimum put $f\text{'}\left(x\right)=0$,

$$\begin{gathered} 6x^2-42x+36=0 \\ \Rightarrow x^2-7x+6=0 \\ \Rightarrow x^2-6x-x+6=0 \\ \Rightarrow \left(x-1\right)\left(x-6\right)=0 \\ \Rightarrow x=1,6 \end{gathered}$$

Since we have two values of $x$ where $f\text{'}\left(x\right)=0$

Differentiate $f\text{'}\left(x\right)$ with respect to $x$,

$f\text{'}\text{'}\left(x\right)=12x-42$

Substitute the value of $x$,

$\begin{array}{rcl}f\text{'}\text{'}\left(1\right)& =& 12-42\\ & =& -30\end{array}$

$\begin{array}{rcl}f\text{'}\text{'}\left(6\right)& =& 12\left(6\right)-42\\ & =& 30\end{array}$

Since $f\text{'}\text{'}\left(x\right)$ is positive at $x=6$,

Therefore function have minimum value at $x=6$.

Now, evaluate the minimum value of the function,

$\begin{array}{rcl}f{\left(x\right)}_{min}& =& 2{\left(6\right)}^3-21{\left(6\right)}^2+36\left(6\right)-20\\ & =& 432-756+216-20\\ & =& -128\end{array}$

Therefore the minimum value of the function $f\left(x\right)=2x^3-21x^2+36x-20$ is $-128$.

Hence option $\left(A\right)$ is the correct answer.