Tutorial on importance of first principle of checking point of maxima and minima

Question

vishweshshrimali5 · Answer

This is basically for the students who have studied calculus and maxima - minima.

vishweshshrimali5 · Answer

We all know there are two methods to check whether a point is a point of maxima or minima or point of inflection.

While many students are aware of the method in which the second derivative is used to identify the nature of the point.

A simple example for this is:

Consider some function say $\large{f(x) = sin^2{x}}$. Now, if I want to evaluate the maximum and minimum value of f(x).

First, you find f'(x). Now,

$\large{f'(x) = 2sin(x)cos(x)}$

Now, lets put it equal to zero to find the critical points. So,

$\large{2sin(x)cos(x) = 0}$

$\large{\implies sin(x) = 0 ~or~ cos(x) = 0}$

Now, lets find out f''(x).

$\large{f''(x) = 2(cos^2(x) - sin^2(x))}$

Now, find out by putting the critical values in this equation and checking the sign of second derivative of f(x) with respect to x to identify whether the point is a point of inflection, minima or maxima.

So,

When $\large{sin(x) = 0}$, 

$\large{f"(n\pi) = 2 >0}$

So, there is minima at $\large{sin(x) = 0}$.

Similarly, on checking for $\large{cos(x) = 0}$, we get,

$\large{f"(\cfrac{(2n+1)\pi}{2}) = -2 < 0}$

So, there is maxima for $\large{cos(x) = 0}$.

Now, on putting the corresponding values of x in f(x) one by one, we get,

$\large{f_{min}(x) = 0}$

$\large{f_{max}(x) = 1}$

vishweshshrimali5 · Answer

As a major portion of students are aware about the above method I decided to cut it short.

vishweshshrimali5 · Answer

But, what if the question would have to be solved by first principle. For this, we only need to find f'(x) and critical values and then find how the sign changes around them.

When f'(x) changes from -ve to +ve, the point is minima.

If it changes from +ve to -ve, the point is maxima.

If the sign does not change, the point is the point of inflection.

vishweshshrimali5 · Answer

For example, we take a new example if 

$$f(x) = \int\limits_{2}^{x}[t(t-3)^3(t-5)^5(t-7)^7(e^t-1)dt]$$ has maxima at x = a, then evaluate:

$$\int\limits_{0}^{a-3}(11x^{10} - 5x^4 - 1)dx$$

vishweshshrimali5 · Answer

You can observe how different people solve a question in different ways.

@ganeshie8 , @mukushla , @Miracrown , @mathmale 

(Can you please give a try to this question? I want to make the students understand the need of having knowledge of alternative methods)

vishweshshrimali5 · Answer

I will be back in half an hour. I have to go and have dinner. Sorry

vishweshshrimali5 · Answer

Hi I am back.

vishweshshrimali5 · Answer

Okay while solving such questions, its necessary to first determine f'(x).

So, lets find out f'(x);

Remember when :

$$\cfrac{d}{dx}\int\limits_{f(x)}^{g(x)}\alpha(x)dx = \alpha(g(x)).g'(x)-\alpha(f(x)).f'(x)$$

vishweshshrimali5 · Answer

Now, 

$$f(x) = \int\limits_{2}^{x}[t(t-3)^3(t-5)^5(t-7)^7(e^t-1)dt]$$

So,

$$f'(x) = x(x-3)^3(x-5)^5(x-7)^7(e^x-1)$$

vishweshshrimali5 · Answer

Now, I can easily find out the critical points which are,

x = 0,3,5,7

Now, if I would have proceeded by using the second principle, I would have got confused and stuck up.

So, I am going to use the first principle,

Now, I observe this:

|dw:1402329823782:dw|

By this figure I mean to convey this that for example, when x was less than 0, f'(x) was -ve and when 1> x > 0, f'(x) remains negative and similarly I can understand the same for other critical points.