Video 1: “Some Function Problem”

Example 1

The graph above shows the graph of y=g(x) if the function h is defined by h(x)=g(2x)+2. What is the value of h(1)?

Information on the graph h(x)=g(2x)+2. We are looking for h(1), the function h(1)=g(2)+2. Based on the graph that at g(2)=1, so we have h(1)=1+2 and final answer h(1)=3.

Example 2

Let the function f be defined by f(x)=x+1 if 2f(p)=20. What is the value of f(3p)?

We looking for f(3p), but first we are looking for what is f when x=3p. we have f(x)=x+1 and similar with 2f(p)=20 equal f(p)=10. Substitute x with p, the function become f(p)=p+1. We have f(p)=10 and the function become f(p)=p+1=10, we get p=1. Substitute p=1 into x=3p, we get x=27, that not final answer. The last, insert x=27 into function f(x)=x+1, so we get f(27)=28, that the final answer.

Example 3

In the xy coordinate plane, the graph of x=y²-4 intersect l at (0,p) and (5,t). what is the greatest possible value of slope of l?

We looking for greatest m. we have x=y²-4 and line l. Line l have slope m which function m=y₂-y₁ over x₂-x_1. The graph of x=y²-4 intersect l at (0,p) and (5,t), to looking for value of p and t, we are substitute (0,p) and (5,t) into x=y²-4, so we get p=2 and t=3. And than, apply (0,2) and (5,3) into m=y₂-y₁ over x₂-x₁, the function become 3-2 over 5-0 or 1 over 5. So, the greatest value of the slope of l is 1 over 5.

Video 2: “Factoring of polynomial”

One way to find factorial of polynomial is to form algebraic long division, for example lets try to see if x-3 a factor of x³-7x-6.

When dividing x-3 into x³-7x-6, first serret the problem long division problem at elementary school. Here is dividing x-3 into x³+0x²-7x-6, the zero is there because no second degree term. Now, you must ask your self what times x give x³, of course it’s x². So you write x² as a part equation and then multiply x-3 by x² which give you x³-3x², which you subtract from x³+0x²-7x-6 to get 3x². Bring it down, -7x, you have x³-7x. Just looking the first term, x goes into 3x², 3x times, 3x is the next part the answer. Multiply x-3 by 3x for predict x²-9x, subtracting and you get 2x and bring down -6, so you get 2x-6. Now see x-3 divide evenly into 2x-6 which equal 2 without remainder. So the solution for the long division problem x³-7x-6 divide by x-3 is x²+3x+2.

Since x-3 divide into x³-7x-6 evenly without remainder, then x-3 is a factor of x³-7x-6. The equation x²+3x+2 is also a factor of x³-7x-6. We know now that x³-7x-6=(x-3)( x²+3x+2). The quadratic expression, x²+3x+2, can be factor into (x+1)(x+2). So, x³-7x-6=(x-3)( x+1)(x+2). Bring the factor x³-7x-6 to zero, we get 0=(x-3)(x+1)(x+2), thus either x-3=0 or x+1=0 or x+2=0. Solving all the equation for x, we get x=3, x=-1, x=-2. The roots of x³-7x-6 are 3, -1, -2.

Now, there 3

Video 3: “Precalculus Graph”

Let’s begin by discussion graph of a rational function which can have discontinuities, why?, because rational function has a polynomial in the denominator.

It possible some value of x will lead division by zero. Let’s if f of x equal x plus 2 over x minus 1, when x=1 the function value become 1+2 over 1-1, which is 3 over 0 with zero in denominator. For this function choosing x=1 is a bad idea. This function will be break in a graph.

Let’s insert 0 into f(x) equal x+2 over x-1, so we get 0+2 over 0-1 which is 2 over -1 or -2. So, you can draw in a graph at (0,-2), and let’s to try insert x=1 this time you get 1+2 over 1-1 which equal 3 over 0. That as we know is impossible, it’s mean you can’t compute value when x=1. In the graph x=1 doesn’t have value.

Rational function don’t always work in this way. Take the graph function f(x) equal 1 over x²+1, not all rational function will give zero in denominator. Don’t forget the general rule, rational function denominator can be zero.

A break can show up in two ways. The simply type break is missing point in the graph. The function y equal x²-x-6 over x-3 will be break when x=3 because the function become zero over zero, that is not possible, not feasible and not allowed. This typical example is missing point syndrome.

When result is zero over zero, that possible to factoring top and button rational function and simplify. In example hand, y=x²-x-6 over x-3, top factor (x-3)(x+2), (x-3) in the top cancel with the button, so the function simplify become y=x+2.

The other one is removable singularity appear simply as missing on a graph and of course when x lead 0 over 0, for this kind the break if you factor and simplify rational function, division by 0 can be avoided.

roots for the 3^rd degree equation. Here remember quadratic (2^nd degree) equation always have at most 2 roots. A 4^th degree equation would have four or fewer roots and so on. The degree of a polynomial equation always limits the number of roots.

Let’s review long division process for a 3^rd order polynomial, first find a partial equation of x² by dividing x into x³ to get x². Then multiply x² by the divisor and subtract the product from the divided. Repeat the process until you either “clear it out” or “reach a remainder”.

Video 4: “Inverse Function and Existence of an inverse”

Some functions do not have inverse functions. For example, consider f(x) = x². Used horizontal test there are two numbers that f same value, example f(2) = 4 and f(-2) = 4. If f had an inverse, then the fact that f(2) = 4 would imply that the inverse of f takes 4 back to 2. On the other hand, since f(-2) = 4, the inverse of f would have to take 4 to -2. Therefore, there is no function that is the inverse of f. The function f(x) = x² have inverse in interval 0<=x, so the function is invertible.

Let start with function y=2x-1. Look at the graph of f is a line with slope 2, so it passes the horizontal line test and does have an inverse.

There are two steps required to evaluate f at a number x. First we multiply x by 2, then we subtract 1. Thinking of the inverse function as undoing what f did, we must undo these steps in reverse order. The steps required to evaluate f^-1 are to first undo the subtracting of 1 by adding 1. Then we undo multiplication by 2 by dividing by 2.

Therefore, x = (y +1)/2 and then interchange y and x. the function become y=(x+1)/2.

We have function f(x)=2x-1 and the other function g(x)=x+1 over 2. If g=f^-1 and substitute g(x) into f(x), so we get f(g(x))=f(f^-1(x)) and substitute f(x) into g(x) and we get g(f(x))=f^-1(f(x))=x.