CALCULUS : CALCULUS Differentiation Integeration
Today we are going to learn a particular
topic in differentiation.i.e How to find
Maximum and minimum value of a function.
Slide 2 : Let y = f(x) be a function.
Now let us know the steps involving to
find Maximum and Minimum Value of the given
function.
Step 1 : Step 1 Find f ’(x) (First Derivative)
Step 2
Find f “(x) (Second derivative) dy
dx d^2y
dx^2
Step 3 : Step 3 Find 0 and find x = a,b
Step 4
sub x = a , x = b in the second derivation
That is dy
dx d^2y
dx^2
Slide 5 : Case (i)
At x= a , If d^2y
dx^2
The Function is Minimum at x=a
The Minimum Value y = f(a)
Slide 6 : Case (i)
At x= b , If d^2y
dx^2
The Function is Maximum at x=b
The Maximum Value y = f(b)
Case (iii) : Case (iii) if when x=a,x=b we can say
There is no Maximum and Minimum value for
the function at x = a and x= b.Let us see the
above procedure for finding the Maximum and
Minimum Value of a function in the following
examples d^2y
dx^2
Ex :1 y = 4x^3 + 2x^2 : Ex :1 y = 4x^3 + 2x^2 Differentiate with respect to ‘x’
4*3x^(3-1) + 2 * 2x^(2-1)
12x^2 + 4x dy
dx dy
dx
Again Differentiate with respect to x : Again Differentiate with respect to x 12*2x^(2-1) + 4*1
24x+4 d^2y
dx^2 d^2y
dx^2
Slide 10 : Now equalize dy
dx 0 12 x^2+4x=0
4x*3x + 4x = 0
4x ( 3x+1) = 0
Either 4x = 0 or 3x+1= 0
4x 0 3x + 1 = 0
4 4 - 1 -1
Therefore x = 0 3x = -1
Slide 11 : 4x 0 3x + 1 = 0
4 4 - 1 -1
3x = -1
x = 0
3x -1
3 3
x = -1
3
Slide 12 : When we equalize dy/dx to 0, we get two values for ‘x’
They are x = 0 and x = -1
3
Now we have to substitue this values in d^2y
dx^2
Slide 13 : At x = 0 d^2y
dx^2
d^2y
dx^2
d^2y
dx^2 = 24(0) +4 = 0 + 4 = 4 > 0
Slide 14 : Therefore at x = 0, the given function
is Minimum.
The Minimum value y = 4(0) +2(0)
y = 0 + 0
y = 0
Slide 15 : At x = -1 d^2y
3 dx^2
d^2y
dx^2
d^2y
dx^2 = 24 (-1/3) +4 = -8+4 = -4 < 0
Slide 16 : There for the given function is Maximum at
x = -1/3.
The Maximum value y = 4 (-1/3)^3 + 2 (-1/3)^2
y = 4(-1/27) + 2(1/9)
y = -4 + 2 -4 + 6
27 9 27 27
y = -4 + 6 2
27 27