Question

The minimum value of the function $y=2{{x}^{3}}-21{{x}^{2}}+36x-20$ is

1. $-128$
2. $-126$
3. $-120$
4. None of these

Answer 1

In this type of question you need to find the first derivative equation of the function and then put that derivative equals to zero, this equation will give you points of local maxima or minima, and then find double derivative at the points you found, the point which give the positive derivative will give you minimum of function.

Complete step by step answer:
First we are going to differentiate one time the given function, because the first derivative of a function shows the nature of the slope of that point. So when this is equal to zero then the rate of the function will be zero means there may be exist a minima or maxima. So, first we find the first derivative.
Let say the function be $f(x)=2{{x}^{3}}-21{{x}^{2}}+36x-20$ .
So, on differentiating with respect to $x$ :
Here, to differentiate we will use the formula:
${{x}^{n}}=n{{x}^{n-1}}$
$\Rightarrow f'(x)=6{{x}^{2}}-42x+36$
Now, we have got the required derivative, now we will put this derivative equal to zero.
Because the derivative provides information about the gradient or slope of the graph of a function we can use it to locate points on a graph where the gradient is zero to find maxima or minima.
Therefore,
$\Rightarrow f'(x)=0$
$\Rightarrow 6{{x}^{2}}-42x+36=0$
On taking common $6$ from above equation we got,
$\Rightarrow {{x}^{2}}-7x+6=0$
$\Rightarrow (x-1)(x-6)=0$
Since we have the equation like $a.b=0$
So, possibilities are $a=0$ or $b=0$
Therefore, on taking one factor equal to zero we got,

& \Rightarrow (x-1)=0 \\\ & \Rightarrow x=1 \\\ \end{aligned}$$ And on taking other factor equals to zero we got, $$\begin{aligned} & \Rightarrow (x-6)=0 \\\ & \Rightarrow x=6 \\\ \end{aligned}$$ Now the points we found can give us maximum or minimum depending on whether the function was accelerating or de-accelerating at that point. In other words if the rate of function becomes zero at a point then it clearly means function may be de-accelerating so that its rate becomes zero or it may be accelerating in negative direction. So to find that rate of slope we will further differentiate the derivative of function. So we have, $$f'(x)=6{{x}^{2}}-42x+36$$ So on differentiating further: $$\Rightarrow f''(x)=12x-42$$ Now checking double derivative for the values we found: $$\begin{aligned} & f''(1)=12(1)-42 \\\ & \Rightarrow f''(1)=-30 \\\ \end{aligned}$$ Since it is negative it will not be considered as a point of minima. Now, $$\begin{aligned} & f''(6)=12(6)-42 \\\ & \Rightarrow f''(6)=30 \\\ \end{aligned}$$ Since it is positive it will be considered as a point of minima. And the minimum value we get is: $$\begin{aligned} & f(x)=2{{x}^{3}}-21{{x}^{2}}+36x-20 \\\ & \Rightarrow f(6)=2{{(6)}^{3}}-21{{(6)}^{2}}+36(6)-20 \\\ & \Rightarrow f(6)=432-756+216-20 \\\ & \Rightarrow f(6)=-128 \\\ \end{aligned}$$ So, the absolute minimum of the given function is $$-128$$ . **So, the correct answer is “Option 1”.** **Note:** Such applications of maxima and minima exist in economics, business, and engineering. Many can be solved using the methods of differential calculus described above. For example, in any manufacturing business it is usually possible to express profit as a function of the number of units sold.