How to find critical points for an equation

In calculus, a critical point of a function of a real variable is any value in the domain where either the function is not differentiable or its derivative is 0. The value of the function at a critical is a critical value of the function. These definitions admit generalizations to functions of several variables, differentiable maps between R^m and Rⁿ, and differentiable maps between differentiable manifolds.

Lets see a problem from calculus help with probability problems,which explains us
more about this .

Question:-

f(x)=3x⁴-4x³-12x²+6

Differentiating with respect to x

Solution:-

f'(x)=12x³-12x²-24x

we know f'(x)=0 ,so

12x³-12x²-24x=0

12x(x²-x-2)=0

12x(x²-2x+x-2)=0

12x(x-2)(x+1) = 0

x(x-2)(x+1) = 0

Now we use the zero product property ,which states that
if ab=0 ,then a=0 and b=0,this property is from geometry terms and definitions .

so x=0 (x-2)=0 (x+1)=0

x=0 x=2 x=-1

Therefore,x=0,-1,2 are the critical points for y=f(x)

Solving a integral problem by using substitution method

Topic:- Integration

Integration is an important concept together with differentiation, forms one of the main operations in calculus help
Given a function ƒ of a real variable x and an interval [a, b] of the real line, the definite integral

is defined informally to be the net signed area of the region in the xy-plane bounded by the graph of ƒ, the x-axis, and the vertical lines x = a and x = b.

Let's work out a simple example on this.
Question:-

solve ∫ ( x / √x²-9 )

Answer:-

We do it by substitution method.
 Let u = x2-9
    
    du
    ---- = 2x
    dx
 
   du
or ---- = xdx
   2
substituting these values ,the integral becomes

∫ (du/2) / √u

= 1/2 ∫ (u)-1/2 du

        (u)-1/2 + 1
= 1/2 ---------------
         -1/2 + 1
by equalizing the denominators
   
    -1+2
   ------- = 1/2
      2
So the integral becomes
    
        (u)1/2
= 1/2 ---------------   + c
         1/2

= (x2-9)1/2+C (as u =x2-9)

= √( x2-9 ) +c  is the Answer

For more help on this ,you can reply me .

Using Shell or Disc Method to Find Volume of the Solid

Disc method and Shell(cylinder) method of integration are the two different methods of finding volume of solid of a revolution, using rectangular coordination system the functions are defined in terms of x in the below problem.

Topic : Disc or Cylinder Method of Finding Volume of the Sphere.
Problem : Use the disc or shell method to find the volume of the solid generated by revolving the regions bounded by the graphs of the equations about the specified line y = 8.
y=x³ y=0, x=2
Solution :
Here the solid is rotated along x-axis
y = x³ => x = (y)^1/3

or x = y^1/3

when y = 8, x = (8)^1/3 = (2³)^1/3 = 2^3*1/3 = 2

So a = 0, b = 2

Volume of a Solid by rotating about y = 8 is given by:

V = 2π₀∫²(y)^1/3 . y dy

= 2π₀∫²(y)^4/3 dy

= 2π[y^7/3/(7/3)₀]²

= 2π[(2)^7/3/(7/3)- (0)^7/3/(7/3)]

= 2π[((2⁷)^1/3/(7/3)]

= 2π ³√(128)/(7/3)

= 6.³√(128)π/7

= 4.32π