Category Archives: Calculus

Solved Problems in Calculus

Solving Quadratic Equations II: Taking Square Roots


A quadratic equation can be solved by taking the square root of both sides of the equation. This method uses the square root property,
y^2=z \to y=\pm \sqrt{z}
Before taking the square root, the equation must be arranged with the x2 term isolated on the left- hand side of the equation and its coefficient reduced to 1. There are four steps in solving quadratic equations by this method:
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Solving Quadratic Equations I: Factoring (Grouping)

Sometimes, quadratic equations can be solved by factorizing, which is also called grouping. The solve by factoring process is usually consists of three major steps:
All terms should be moved to one side of the equation using addition or subtraction. Rewrite the equation so that the left side of the equation is set equal to 0. For example, if the original equation is  x^2 = 5x - 3 , it should be rewritten as  x^2 - 5x + 3 = 0 .
The equation should be factored completely. It will have two factors.
Each factor should be set equal to zero. Since factors are of first power, they are easy to be solved.
All of the solutions should be combined to obtain the full solution set for the original equation.
The catch of this method is the finding of factors. Except some trivial cases, factoring quadratic equations is not easier than using other methods to solve the quadratic equations. Here are a few tips to help you factorize a quadratic equations:

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Solving Equations II: Radical Equations


A radical equation is an equation in which a variable appears under a radical sign. It may also have more than one radical. Let's see some examples of radical equations:

\sqrt{x-1}=2

\sqrt{2x+4}=\sqrt{5}

\sqrt{4x+1}+2x=-1

\sqrt{x+1}=\sqrt{2x+1}+10

\sqrt{10-x}+\sqrt{x-5}=3

\sqrt{x^2+3}=3

\sqrt[3]{2x-4}=-4

\sqrt{\sqrt{x+2}+7}=4

\sqrt[3]{x^2-1}=1

\sqrt[3]{6x^2+2}-1=x

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Solving Equations I: Linear Equations


In a linear equation, each term is either a constant or the product of a constant and a single variable of degree 1. It can have one or more variables. Here are some linear equations:

x+1=0
2x+4=6
-3x+1=x+4
1+5x=0
2-3x=3+2x
\frac{1}{2}x+\frac{3}{4}=\frac{1}{3}
\frac{x-1}{3}+\frac{1}{4}=\frac{5}{2}
\frac{x}{c}+b=a where  c\neq 0 is a constant
\sqrt{3}x-\sqrt{2}=\sqrt{5}

However, the following equations are not linear:
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Problem 2-10: Diffrentiating


Diffrentiate
a)  f_1\left(x\right)= \displaystyle\frac{1-2x^2}{1+x+x^2}
b)  f_2\left(x\right)= (1-2x^2)(1+x+x^2)

Solution
The following rules can be used:
I)  \displaystyle\frac{d}{dx}ax^n=anx^{n-1}
II)  \displaystyle\frac{d}{dx}\left(u+v\right)=\frac{du}{dx}+\frac{dv}{dx}

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Problem 2-9: Differentiating Polynomial and Rational Functions


Differentiate
a)  f\left(x\right)= 1+ 2x
b)  f\left(x\right)= 1 - 2x + 2x^2
c)  f\left(x\right)= 1- x^2 +\frac{1}{x}
d)  f\left(x\right)= 1 - x +\frac{1}{1-x}


Solution
The following rules can be used:
I)  \displaystyle\frac{d}{dx}ax^n=anx^{n-1}
II)  \displaystyle\frac{d}{dx}\left(u+v\right)=\frac{du}{dx}+\frac{dv}{dx}
III)  \displaystyle\frac{d}{dx}\frac{u}{v}=\frac{1}{v^2}\left(v\frac{du}{dx}-u\frac{dv}{dx}\right)
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