# The Science That Draws Necessary Conclusions

To Learn Math

## Taking a Look at Wolfram|Alpha

Lets take a small break from learning and look at a somewhat new tool. Wolfram|Alpha is a “computational knowledge engine that draws on multiple sources to answer user queries directly.” While this tool has numerous applications, I will take a look at how it can be used for math.

Simply by entering a number, you will gain access to a variety of information about it. It includes some information about the number in different forms such as a Roman numeral or binary. More useful information may be found in its prime factors or its properties. Other information is included which would be useful to some individuals.

Now, onto a more interesting application. Wolfram|Alpha accepts functions as input. The first thing noticed is that Wolfram|Alpha will graph the function. Then it will provide you with the figure of the base function. With the inclusion of the roots of the graph, this basic information may all that is needed.

The other information returned is what I find to be the most valuable. Not only does Wolfram|Alpha take the derivative of the function as well as the integral, but it also will show the steps it took to reach its answer.

Evidently, Wolfram|Alpha has some potential. I encourage you to tinker with different functions in it. Also, try putting some of the functions on this site into the engine. For many more examples visit this link.

Written by todizzle91

August 24th, 2009 at 9:26 am

Posted in not math

## Shortcuts for Finding Derivatives

Using the derivative formula to find the derivative of a function is a hassle. There are many shortcuts to finding the derivative. The first one is the Power Rule. It can be used with functions which are equal to a polynomial like $x^3 + 5x^2 - 7$. Each term’s coefficient is multiplied by the exponent and the exponent decreases by one. Let’s look at $f(x)=3x^2$. The derivative’s coefficient will be 6 because the coefficient of the term, 3, is multiplied by the exponent, 2. The exponent will be 1 because the exponent is decreased by one. Therefore, the derivative is $f'(x)=6x$. If the function has more than one term, use the same procedure for each term. Now let’s use the power rule to find the derivative of $g(x)=2x^4-5x^2+3x-2$. The derivative is $g'(x)=8x^3-10x+3$. It is important to notice that the constant, -2, is eliminated since x has an exponent of zero. Also, 3x becomes 3 since anything raised to the power of 0 is equal to 1.

Another useful shortcut is the Product Rule. It is useful when the definition is two polynomials multiplied by each other but cannot easily be simplified such as $f(x) = (x^2 - 2)(3x^2 + x - 4)$. If u is assigned to the first polynomial and v is assigned to the second polynomial, the derivative is $u\frac{dv}{dx}+v\frac{du}{dx}$. So in $f(x)$ $u = x^2 - 2$ and $v = 3x^2 + x - 4$. Using the power rule, $\frac{du}{dx} = 2x$ and $\frac{dv}{dx} = 6x + 1$. Therefore, by using the product rule $f'(x) = (x^2 - 2)(6x + 1) + (3x^2 + x - 4)(2x) = 12x^3 + 3x^2 - 20x - 2$.

The last rule I am going to discuss for now is the Quotient Rule. The Quotient Rule is used when a polynomial is being divided by another polynomial. Where $y=\frac{u}{v}$, $\frac{dy}{dx} = \frac{ v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$. For example, let $f(x)=\frac{x^2+2}{x-1}$. The u is $x^2 + 2$ and v is $x - 1$. Once again utilizing the power rule, we find that $\frac{du}{dx} = 2x$ and $\frac{dv}{dx} = 1$. Plugging these values into the quotient rule yeilds $f'(x) = \frac{(x-1)(2x)-(x^2+2)(1)}{(x-1)^2} = \frac{x^2-2x-2}{x^2-2x+1}$.

Proper use of these rules save time. On some occasions, multiple rules will need to be applied. There is another rule I have not yet mentioned called the chain rule. It makes taking the derivative of a function within a function possible. That post will be saved for after the explanation of the usefulness of derivatives.

Written by todizzle91

August 13th, 2009 at 10:35 am

Posted in Calculus

## Intro to Derivatives

with one comment

The derivative of f(x) at point a is the slope of the tangent line at point a or $\displaystyle f'(x)=\lim_{h \to 0}\frac{f(x+h) - f(x)}{h}$. Allow me to explain.

As seen in the picture above, a secant line is a line that passes through two points of a function and a tangent line touches only one. The function to find the slope of a secant line that passes through is $m=\frac{f(x+h)-f(x)}{h}$ where h is the change in x or $\Delta x$. This is simply the slope formula, $m=\frac{y_2-y_1}{x_2-x_1}$, modified. Taking the limit as h approaches 0 means that the $\Delta x$ would “become zero”, so the line would only pass through that one point. That is a tangent line.

The picture above should help make it clearer. It illustrates using the derivative formula to find the slope of the tangent line at x=1 on $f(x) = -x^2 + 5$. The red line is the secant line as h approaches zero from the left (values of h are -3, -2, and -1). The blue line is the secant line as h approaches zero from the right (values of h are 3 and 2). The purple line is the tangent line; it touches the function at only one point. Now I will show you how it is done algebraically. I have written it out since I believe that is easier to understand.

The first thing I do is replace f(x) with its value $-x^2 + 5$ replacing x with x + h in the first term. I then multiply everything out so I am left with a polynomial. The $x^2$‘s and 5‘s cancel out. Now it is possible to cancel out an h in each term. After doing so, I plugged in zero for h to leave me with the derivative -2x.

The derivative only tells you the slope of the tangent line, not the equation. Since you know the slope of the line and a point on the line, it is not difficult to find the equation of the tangent line, g(x) in this case.

There are many tricks and shortcuts to finding the derivative of a line. The derivative formula is usually the most difficult. Things to remember: The derivative of $f(x)$ is $f'(x)$. A derivative of the derivative yields the second derivative or $f''(x)$. The derivative of y is written as $\frac{dy}{dx}$. In the next two posts I will delve into some of the shortcuts to finding the derivative of a function and why derivatives are useful.

Written by todizzle91

February 16th, 2009 at 12:20 pm

Posted in Calculus

## Algebraically Finding a Limit

Finding a limit may be as simple as plugging a number into a function, or it may require manipulating the function if it is undefined at that point. $f(x) = 3x +1$ is a linear function continuous at all points. The limit as x approaches any point a is f(a), $\displaystyle \lim_{x \to a}f(x) = f(a)$. To find the limit as x approaches 2 in f(x), just plug x into f(x), $\displaystyle \lim_{x \to 2}f(x)=f(2)=3(2)+1=7$.

Now look at the rational function $g(x)=\frac{x+1}{x-1}$. g(x) is continuous at every point except when x=1. In order to solve the limit as x approaches 1, g(x) must be manipulated to defined at 1. Multiplying g(x) by $\frac{x-1}{x-1}$ will return the exact same function with the exception of g(1) being defined, (x+1)(x-1). The limit can be taken using this function so $\displaystyle \lim_{x \to 1}g(x)=\lim_{x \to 1}(x+1)(x-1)=0$.

Another good example is $\displaystyle \lim_{x \to \pm 2}h(x)$ if $h(x)=\frac{x+2}{x^2-4}$. h(x) is undefined at -2 and 2 so the function needs to be manipulated to be defined at those two points. $x^2-4$ is the difference of two squares so it can be factored to $(x-2)(x+2)$. The numerator and denominator both share the factor (x+2) so it can be cancelled leaving the function $\frac{1}{x-2}$. Now it is clear that $\displaystyle \lim_{x \to -2}h(x)=\lim_{x \to -2}\frac{1}{x-2}=-\frac{1}{4}$; however the new function is still undefined at 2. Because the function cannot be simplified any more, the limit as x approaches 2 does not exits, $\displaystyle \lim_{x \to 2}h(x) = DNE$.

Written by todizzle91

December 5th, 2008 at 10:36 am

Posted in Calculus

## Continuity and Discontinuity

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A function is said to be continuous at a if $\displaystyle \lim_{x \to a} f(x) = f(a)$. A continuous function’s points are all connected. Below are examples of functions that are not continuous.

The graph above is an example of jump discontinuity. There is jump continuity at each integer. The limit from the positive and negative direction of each is a finite number, but they are not equal. Let’s look at the point where $x=2$. $\displaystyle \lim_{x \to 2^-}y=1$ and $\displaystyle \lim_{x \to 2^+}y=2$. Because both numbers are not infinite and are not equal, there is jump discontinuity where $x=2$.

The graph above is an example of infinite discontinuity. This is when the limit of a point from the negative or positive side is infinite or does not exist. There is infinite discontinuity at the point where $x=0$. $\displaystyle \lim_{x \to 0^-}y=\infty$ and $\displaystyle \lim_{x \to 0^+}y=\infty$. Even though the limit from both sides equal infinity, the two lines never actually meet.

Above is an example of removable discontinuity. Removable discontinuity is when the limit from both sides of a point are equal and finite, but the function is undefined at this point. There is a hole in the graph where $x=-1$. $\displaystyle \lim_{x \to -1^-}y=-2$ and $\displaystyle \lim_{x to -1^+}y=-2$. Even though the limits are equal where $x=-1$, there is no point there.

Written by todizzle91

October 2nd, 2008 at 9:49 pm

Posted in Calculus

## Intro to Limits

A limit is what f(x) approaches as x approaches a value from both directions. The limit of a function f(x) as x approaches 0 is written as $\displaystyle \lim_{x \to 0}f(x)$.

f(x) is pictured above. It is evident that $\displaystyle \lim_{x \to 3}f(x)=5$. As x gets closer to the value 3, f(x) gets closer to 5.

f(x) never needs to equal the number x approaches. $\displaystyle \lim_{x \to 4}g(x)=3$ even though $g(4)=5$.

If there is a vertical asymptote asymptote at x=0, $\displaystyle \lim_{x \to 0}f(x)=DNE$. DNE stands for “does not exist”. The line approaching x from the negative side goes to $-\infty$. From the positive side, the line appproaches $\infty$. Because they are not approaching the same point, the limit is said not to exist.

Limits can  pertain to only the line coming from only one side. It is written as $\displaystyle \lim_{x \to 0^+}f(x)$ if it only pertains to parts of the line where $x>0$ or as $\displaystyle \lim_{x \to 0^-}f(x)$ if it only pertains to parts of the line where $x<0$. If $f(x)=\frac{1}{x}$, then $\displaystyle \lim_{x \to 0^+}f(x) = +\infty$ and $\displaystyle \lim_{x \to 0^-}f(x) = -\infty$.

Before we delve deeper into limits, I will write a post about continuity.

Written by todizzle91

September 26th, 2008 at 3:30 pm

Posted in Calculus

Quadratic equations are equations that have the variable raised to the second power. They are in the form of $f(x) = ax^2 + bx + c$.

The graph of a quadratic equation looks like a cross between a U and a V. There are two roots (where the funtion returns 0 also called a zero); they can be imaginary and both can be the same. In the graph above, the roots are 0.

A common way to find the roots of a quadratic equation is with the quadratic formula. It is $x = \frac{-b \pm \sqrt {b^2-4ac}}{2a}$ where x is the root. It has two solutions, one using the plus sign and the other using the subtraction sign. The nature of the roots can be found with the discriminant. The discriminant is the part from under the square root, $b^2 - 4ac$. If it is negative, there are two imaginary zeros. If the discriminant is 0, there is one real roots. If the discriminant is positive, there are two real roots.

There is more too the quadratic function, but it can be explained along with all other polynomial equations.

Written by todizzle91

June 20th, 2008 at 10:37 am

Posted in Functions

## Linear Functions

A linear function is a function whose graph is a line. They are written in three forms: slope-intercept form, point-slope form, and general form.

Slope-intercept is a common way to write a linear function. Its form is $f(x) = mx+b$. The graph of $f(x) = x$ is shown in the image above. The variables m and b modify this graph.

m is the slope of the line (how steep the line is). In $f(x) = x$, m is equal to 1. As a result, one point can be reached from another point by moving up 1 unit and to the right 1 unit. The function $f(x) = 2x$ has a slope of 2 so it will move up 2 units and to the right 1 unit. The function $f(x) = \frac{1}{2}x$ has a slope of $\frac{1}{2}$ so it will move up 1 unit and to the right 2 units. When m is negative, the line will be moving down rather than up.

b is the y-intercept of the line (where the line intersects the y-axis). The y-intercept of $f(x) = x$ is 0. As a result, the line crosses the y-axis at the origin. The line $f(x) = x - 3$ crosses the y-axis at -3.

Point-slope form is $y - y_1 = m(x - x_1)$. This can be written as a function as $f(x) = m(x - x_1) + y_1$ (y become f(x)). $x_1$ and $y_1$ are the coordinates of a point on the graph. This can be useful if the slope and a point on the graph is known. For example, a line has a slope of 3 and passes through the point (3, 5). $f(x) = m(x - x_1) + y_1$ with the point and slope plugged in becomes $f(x) = 3(x - 3) + 5$. After distributing the 3 and adding the -9 to the 5 gives the graphable slope-point form $f(x) = 3x - 4$.

General form is $Ax + By + C = 0$. It is less commonly used and at the moment not that important. It can be converted to point-slope form by changing it to $f(x) = (-\frac{A}{B})x + (-\frac{C}{B})$. The slope is $-\frac{A}{B}$ and the y-intercept is $-\frac{C}{B}$.

The graph above can be written as $f(x) = 3x - 4$ (slope-point), $f(x) = 3(x - 1) - 1$ (point-slope), or as $y - 3x + 4 = 0$ (general form).

Written by todizzle91

April 23rd, 2008 at 4:59 pm

Posted in Functions

## Introduction to Functions

A relation is composed of ordered pairs of numbers. The set of the first numbers is the domain and the set of the second values is the range of the relation. The first number is called the abscissa and the second number is called the ordinate.

The ordered pairs can be graphed using the Cartesian coordinate system. There are two axes in the coordinate system, the x and y axis. The x-axis is horizontal and the y-axis is vertical. The point where the lines cross is called the origin. That point is (0, 0). The further a point is toward the right, the larger the abscissa is. Moving toward the left from the origin goes negative. The further the point is upward, the larger the ordinate of the point is. The abscissa can be called the x value and the ordinate can be called the y value.

A function is a special relation where each abscissa has no more than one ordinate. This means that any input will always yield the same output. Note that different inputs may share the same output. The input is the independent variable x and the output is the dependent variable y.

The vertical line test tests whether a relation is a function. If a vertical line can be placed on a graph and it crosses through two or more points, then the line is not a function. The graph of a circle is not a function because a vertical line can be drawn that passes through more than one point. The picture of the graph on the right is a function because no matter where a vertical line is placed, it will only pass through one point. Each x value has only one y value.

A functions is written as $f(x) = expression$. f(x) is the same as the y coordinate. So if $f(x) = x^2$, then $f(3) = 9$ and (3, 9) is a point of the graph of f(x). There are numerous functions to learn, but luckily there are patterns.

Written by todizzle91

April 16th, 2008 at 9:29 pm

Posted in Functions