Precalculus Problem Solution

Calculus is one of the learning about the rates of change and measurement of changing quantities in which it is using symbolic notations in the calculations. In precalculus it identifies the values of the function and it does not involved in the problems of calculus but it explores topics that will be applied in calculus. So that precalculus one of the main division in the calculus

Calculus two types, they are.          

• Differential calculus
• Integral calculus       

      Precalculus functions that are given is, Domain and Range Composition, and Difference of Quotients

Differential calculus:

        Differentiations are been used to determine the rates of change of a function and the larger and smaller values of a function.

        It is a branch of Calculus that deals with derivatives and their applications.

Integral Calculus:

         Integration is the branch of the calculus in which it deals with integrals and its application,  in determining areas, equations of curves, or volumes.                                                                                                                

                          

Some precalculus operations:

• The slope of line passing through points of` (x_1, y_1) ` and` (x_2, y_2)` which is given by,

                        m = `(y2 - y1) / (x2 - x1)`

• slope m has the equations. in which passes through the line through the point (x₁, y₁).

                    y - y₁ = m(x - x₁)

• y-intercept the b has the equation of line y = mx + b.It is the line passes through the slope
• Two lines of the equation with slopes m₁ and m₂.

            If the line of the slope is parallel m₁ = m₂ and

            If the line of the slope is perpendicular m₁m₂ = -1.
                                                      

Precalculus problem solution - Example problems:

Precalculus problem solution - problem 1:

Solve the equation with the  curves y = mx, where, m is arbitrary constant.

Solution:

      We have the equation of the curve

               y = mx ... (1)

     Differentiating either side of equation (1) with respect to x, we get

      `dy / dx` = m

     Substitute the value of m in equation (1) we get

               y = ` (dy / dx)` * x

     or 

            x `( dy / dx ) ` - y = 0

    Hence this is the required differential equation.         
Precalculus problem solution - problem 2:
Find the derivatives for,   (i) Y = `4 / x^3`   (ii) Y = 6`sqrt (x^3)`
   Solution :
                  (i). Y = `4 /x^3 `
                        Y = 4 X ^-3
                     ` dy / dx` = (4-3)x^(-3-1)
    Answer     = `(1 / x^(-4))`
               (ii) Y = ` 6sqrt(x^3)`
                      Y =  `6 x^(1 /3)`
             `dy / dx` = `6 (1/3)` `x ^-(2 / 3)`
           `dy / dx ` = `(6/3)x^-(2/3)`
     Answer = `(6/3)x^-(2/3)`