Lecture 16

     

 MATH 121 (Wg Cdr Monir) 

Maxima and Minima

    Some Definitions:
    ♦ Increasing Function: 
If  $y = f(x)$ is a function of x and y increases as x increases in a certain interval, Then y is called an increasing function of x in that interval.

  $$\begin{align*} \frac{dy}{dx} = & \tan \Psi >0 \\ & \Psi < 90^\circ \end{align*}$$  

    ♦ Decreasing Function:
If $y = f(x)$ is a function of x and y decreases as x increases in a certain interval, then y is called a decreasing function of x in that interval.

  $$\begin{align*} \frac{dy}{dx} = & \tan \Psi <0 \\ & \Psi > 90^\circ \end{align*}$$  

    ♦ Concavity:  
If the graph of a function $f$ lies above all of its tangents on an interval l, then it is said to be concave up on l.
If the graph of a function $f$ lies below all of its tangent on an interval l, then it is said to be concave down.

    ♦ Test for concavity
• The function $f$, is concave up on any open interval l where $f''(x)>0$ and it is concave down where $f''(x)<0$ 
• Inflection Point: If $x = a$ inflection point, when
        (1)  $f''(x)>0$ when $x<a$ and $f''(x)<0$ when $x>a$ 
    or
        (2) $f''(x)<0$ when $x<a$ and $f''(x)>0$ when $x>a$.

    Note: Inflection point does not occur where $f''(x) = 0$. The inflection point occurs where the sign of the second derivative change.

Question: Find the maximum and minimum value of the function  $x^5 - 5x^4 + 5 x^3 - 1$. Also, find the inflection point if any.
Solution: Let,  $$\begin{align*} & f(x) = x^5 - 5x^4 + 5x^3 - 1 \\ & f'(x) = 5x^4 - 20x^3 + 15x^2 \end{align*}$$  

For maximum or minimum $f'(x) = 0$
  $$\begin{align*} & 5x^4 - 20x^3 + 15x^2 = 0\\ or, ~ & 5x^2(x^2 - 4x + 3 ) = 0 \\ or, ~ & 5x^2(x - 1) (x - 3) = 0 \\ or, ~ & x = 0, 1,3 \end{align*}$$  
Now,  $f''(x) = 20x^3 - 60x^2 + 30x$ 
For x = 1
  $f''(x)= 20 - 60 +30 = -10 <0$
then the function $f(x)$ is maximum at x = 1 and the maximum value is,                        $$\begin{align*} f(x) = & 1^5 - 5 \times 1^4 + 5 \times 1^3 - 1\\ = & 0 \end{align*}$$                         

For x = 3
  $f''(x) = 20 \times 3^3 - 60 \times 3^3 + 30 \times 3 = 90 >0$  
then the function $f(x)$ is minimum at x = 3 and the minimum value is,  $$\begin{align*} f(x) = & x^5 - 5\times 3^4 + 5 \times 3^3 - 1 \\ = & - 28 \end{align*}$$  

For x = 0
  $$\begin{align*} f'' (x) & = 20 \times 0 - 60 \times 0 + 30 \times 0 \\ & = 0 \\ f'' (x) & = 20 x^3 - 60 x pe 2 + 30x \\ f''' (x) & = 60x^3 - 120x + 30 \\ \textrm{Now for}~~~ & x = 0, f'''(x) = 30 \ne 0 \end{align*}$$  
the function has neither maximum nor minimum at x = 0
At x = 0 is an inflection point