Advanced Functions
Saturday, November 5, 2011
Laws of Logarithms and Change of Base
Logarithm study has a few formulas that are important and key to solving math questions. By remembering them, you will be in line to solve logarithmic problems and, maybe, fast too. There are a few logarithms laws that are formed by combining two logs of the same base.
- Product rule : log(ab) = log(a) + log(b)
- Quotient rule : log(a/b) = log(a) - log(b)
- Power rule : log(xn) = nlog(x) Ex: log39² = 2log39
If the logarithm to the base a is known, then the logarithm to the base b can be obtained by the base change relationship:
Thursday, November 3, 2011
Transformations of Logarithmic Functions
Transformations apply to logarithmic functions the same way as they do to other functions. Recall what we have learnt in previous chapters on how transformations effect a function and its graph.
- f(x) → f(x) + c
- f(x) → f(x - d)
- f(x) → af(x)
- f(x) → f(kx)
The same techniques is applied to logarithmic functions. An equation of a transformed logarithmic function has a form of f(x) = a logB [k (x+d)] + c where a, k, d and c are coefficients while B is the base of the log.
We must first apply the horizontal/vertical stretches or compressions first if any.
- y = log (kx) : Horizontal compression/stretch by factor of |1/k|
- y = a log x : Vertical stretch/compression by a factor of |a|
The next step is to apply any reflections. If a <0, reflect in the x-axis. If k<0, reflect in the y-axis.
Then apply the horizontal or vertical translations to the function.
- y = f(x − d) : shift to the right by d units if d>0; translate left by d units if d<0
- y = log x + c : translate up by c units if c>0; shifts down by c units if c<0
Wednesday, November 2, 2011
Logarithms
Logarithmic functions are the inverse of exponential functions. For example, the inverse of y = ax is y = logax, which is the same as x = ay. For example, if (0, 1) is a point on the graph of an exponential function, then (1,0) would be the corresponding point on the graph of the inverse logarithmic function. This is proven in the graph below.
The graph of a logarithm function is a reflection of its exponential graph under the line y=x.
- The domain is all real numbers and x>0.
- The range is all real numbers with no restrictions.
- has an x-intercept of 1 as it will always cross the x-axis a 1. The graph will never cross the y-axis because x can never equal 0.
- has a vertical asymptote at x=0 as the graph passes through the point (1,0)
Therefore, logarithm equations can be written in exponential form and vice versa. Sometimes, by changing everything in logarithmic form to exponential form, you can solve equations containing logarithms. The little chart below will hopefully help us remember how to convert from one to the other.
Friday, October 21, 2011
Exponential Function
In chapter 6, we leave trigonometric functions (I can't be the only one glad to do so, right?) and move on to exponential functions. The simplest exponential function is in the form of y = ax, a > 0, a ≠ 1. There is a restrictions because if a≤0, you may not get a real number when you substitute it into the equation.
Let's study the function y = 2x and its graph. The base is 2 while the power x is the variable. After evaluating the function and determining the values, a graph of the exponential function can be drawn as shown.
Here are some properties of the exponential function when the base is greater than 1.
- The domain is all real numbers
- The range is all real numbers and y>0.
- has y-intercept of 1 as the graph passes through the point (0,1)
- has a horizontal asymptote at y=0
- The graph is increasing as a > 1
Friday, August 26, 2011
Polynomial Equations
Solving a polynomial equation can be done algebraically if it is factorable or graphically if it not. Note that a polynomial function is y = P(x) and a polynomial equation is when P(x) = 0.
If the polynomial equation is factorable, the roots are determined by solving each factor when it is equal to zero. For example,
The x-intercepts of the graph are 0 and 4/7. The x-intercepts of a polynomial function graph correspond to the real roots of the polynomial equation P(x) = 0. Therefore, the roots for the polynomial equation in the example above is also 0 and 4/7.
P(x) can also be solved using the factor theorem. To solve the cubic x3 - x2 + x -1 = 0,
we first figure out the values that should be tested using the rational zero theorem. If P(x) is a polynomial function with integer coefficients and x = b/a is a zero of P(x), then b is a factor of the constant term while a is a factor of the leading coefficient. Furthermore, ax - b is a factor of P(x). Back to the equation, since a=1 and b=-1, the only possible values of b/a is 1. By substitution, we see that x=1 is a root thus (x-1) is a factor. We then find the other factor by dividing with (x-1) which produces the quotient x2 +1. Thus,
Since a square root of a negative value is not a real number, we do not get a real root from x2 + 1 as it is a complex root. So the only real root is x=1. If a polynomial equation has real coefficients, then either all roots are real or there are an even number of non-real complex roots, in pairs.
On the other hand, if the polynomial equation is not factorable, the roots can be determined from the graph by examining the x-intercepts. By graphing the function using a graphing calculator, you can get a sense of where the roots are and how many real roots exist when it crosses or touches the x-axis.
A set of functions with the same characteristics is a family of functions. When talking about a family of a polynomial function, we are talking about the general equation of the function. Polynomial functions that have the same x-intercepts belong to the same family.
If the polynomial equation is factorable, the roots are determined by solving each factor when it is equal to zero. For example,
The x-intercepts of the graph are 0 and 4/7. The x-intercepts of a polynomial function graph correspond to the real roots of the polynomial equation P(x) = 0. Therefore, the roots for the polynomial equation in the example above is also 0 and 4/7.
P(x) can also be solved using the factor theorem. To solve the cubic x3 - x2 + x -1 = 0,
we first figure out the values that should be tested using the rational zero theorem. If P(x) is a polynomial function with integer coefficients and x = b/a is a zero of P(x), then b is a factor of the constant term while a is a factor of the leading coefficient. Furthermore, ax - b is a factor of P(x). Back to the equation, since a=1 and b=-1, the only possible values of b/a is 1. By substitution, we see that x=1 is a root thus (x-1) is a factor. We then find the other factor by dividing with (x-1) which produces the quotient x2 +1. Thus,
x3 - x2 + x - 1 = (x - 1)(x2 + 1) = 0
x - 1 = 0
x2 + 1 = 0
On the other hand, if the polynomial equation is not factorable, the roots can be determined from the graph by examining the x-intercepts. By graphing the function using a graphing calculator, you can get a sense of where the roots are and how many real roots exist when it crosses or touches the x-axis.
A set of functions with the same characteristics is a family of functions. When talking about a family of a polynomial function, we are talking about the general equation of the function. Polynomial functions that have the same x-intercepts belong to the same family.
Friday, August 19, 2011
Remainder Theorem and Factor Theorem
We started Chapter 2 with long division. Easy, right? Well this time we learned dividing a polynomial function with a binomial. In the following example, P(x) is divided by the factor x-4.
The quotient for the example above is x2 + 4x + 9 with a remainder of 30. Thus, x3 – 7x – 6 = (x – 4) (x2 + 4x + 9) + 30. When dividing, just remember to do so carefully to avoid mistakes.
The Remainder Theorem is an easier way to find the remainder. Instead of using the long division method, take x=b from the factor x-b and substitute it in the polynomial function. From the example above, b=4 so we solve P(4) to find the remainder.
P(4) = (4 – 4)((4)2 + 4(4) + 9) + 30
= (0)(16 + 16 + 9) + 30 = 0 + 30
= 30
Therefore, the remainder theorem:
When you divide a polynomial P(x) by x-b the remainder R will be P(b).
Now, if the remainder is 0, then x-b is a factor of the polynomial. Just as with the Remainder Theorem, the point here is not to do the long division of a given polynomial by a given factor. The same method is used as the Remainder Theorem simply by substituting b to find P(b). The Factor Theorem is just a result of the Remainder Theorem. It's purpose is to check for a zero remainder to determine the factors of the polynomial.
And so we have:
When P(b)=0 then x-b is a factor of the polynomial P(x).
Knowing that x-b is a factor of P(x) is useful as it tells you that b is an x-intercept of the function.
If you are still confused, the video below explains the difference between the Remainder Theorem and the Factor Theorem.
In addition to few examples of both the Remainder Theorem and Factor Theorem, this websites provides an exercise for you to work on. Yes, answers are given. I hope this post has helped you in a way. Thanks for reading. :)
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