Math questions, please help me? :(?

Find the composite functions f o g and g o f for the given functions.
f(x) = 10^x and g(x) = log x

State the domain and range for each:
f(x) = 4x + 1 and g(x) = x^2
f(x) = sin x and g(x) = x^2 - x + 1
f(x) = 10^x and g(x) = log x

If f = {(-2,-1), (-1,2), (0,6), (1,3), (2,-1), (3,-5), (4,0), (5,8)} and g = {(-2,6), (-1,5), (0,-8), (1,3), (2,9), (3,9), (4,0), (5,-2)}, then find f o g and g o f.

and how on earth would I figure this out?
By graphing the functions, determine the region where log2(x+3) > 2x.

(f◦g)(x) = f(g(x))

f(x) = 10^x, so
f(g) = 10^g = 10^(logx) = x
Similarly, g◦f = x
Note that 10^x and logx are inverse functions

domain for all polynomials is all reals
g(x) = x^2 has range y>=0
g(x) = x^2-x+1 = (x-1/2)^2 + 3/4 has range y >= 3/4
since 10^x and logx are inverses, and the range of 10^x is y>0, the domain of logx is x>0

When given actual pairs, just pug and chug

(f◦g)(-2) = f(g(-2)) = f(6) is undefined
(g◦f)(-2) = g(f(-2)) = g(-1) = 5
and so on

Visit wolframalpha.com and enter

plot y=log2(x+3) and y=2x

to see the graphs. Note that the lower solution is not exactly -3, since log(0) is not defined.

Sure! I can help you with your math questions. Let's tackle them one by one.

1. Find the composite functions f o g and g o f for the given functions:
To find f o g, you need to substitute g(x) into f(x). So, f o g(x) = f(g(x)).
f(g(x)) = f(log x) = 10^(log x).
Remember that 10^(log x) cancels out, leaving you with just x. Therefore, f o g(x) = x.

To find g o f, you need to substitute f(x) into g(x). So, g o f(x) = g(f(x)).
g(f(x)) = g(10^x) = log(10^x) = x.
Similarly, log(10^x) cancels out, leaving you with just x. Hence, g o f(x) = x.

2. State the domain and range for each:
a) f(x) = 4x + 1 and g(x) = x^2.
For f(x) = 4x + 1, the domain is all real numbers since there are no restrictions on x. The range is also all real numbers since the function is a linear equation.

For g(x) = x^2, the domain is all real numbers because squaring any real number will give a real number. The range is all non-negative real numbers since the square of a real number is always non-negative.

b) f(x) = sin x and g(x) = x^2 - x + 1.
For f(x) = sin x, the domain is all real numbers since sine is defined for any value of x. The range is between -1 and 1, inclusive, since the sine function oscillates between these values.

For g(x) = x^2 - x + 1, the domain is all real numbers (no restrictions on x). The range is all real numbers greater than or equal to 3/4, which is the vertex of the quadratic function.

c) f(x) = 10^x and g(x) = log x.
For f(x) = 10^x, the domain is all real numbers since raising 10 to any real power is defined. The range is all positive real numbers as 10^x will never be zero or negative.

For g(x) = log x, the domain is all positive real numbers since logarithms are not defined for zero or negative values. The range is all real numbers since logarithms can take on any real value.

3. If f = {(-2,-1), (-1,2), (0,6), (1,3), (2,-1), (3,-5), (4,0), (5,8)} and g = {(-2,6), (-1,5), (0,-8), (1,3), (2,9), (3,9), (4,0), (5,-2)}, then to find f o g and g o f:
To find f o g, substitute the y-values of g into f. So, for each (x, y) in g, we substitute y into f.
f o g = {(-2, f(g(-2))), (-1, f(g(-1))), (0, f(g(0))), (1, f(g(1))), (2, f(g(2))), (3, f(g(3))), (4, f(g(4))), (5, f(g(5)))}.
Substitute the values, and you will get f o g.

To find g o f, substitute the y-values of f into g. So, for each (x, y) in f, we substitute y into g.
g o f = {(-2, g(f(-2))), (-1, g(f(-1))), (0, g(f(0))), (1, g(f(1))), (2, g(f(2))), (3, g(f(3))), (4, g(f(4))), (5, g(f(5)))}.
Again, substitute the values, and you will get g o f.

4. By graphing the functions, determine the region where log2(x+3) > 2x:
To solve the inequality log2(x+3) > 2x, we can start by graphing the two functions y = log2(x+3) and y = 2x. Then we find the region where the graph of log2(x+3) is above the graph of 2x.

To graph y = log2(x+3), you can start with a table of values and plot the points. Remember that log2(x+3) is only defined for x > -3. Draw the graph accordingly.

To graph y = 2x, you can use the slope-intercept form, which is already in this form. Plot a few points and draw the line.

The region where log2(x+3) is greater than 2x is where the graph of log2(x+3) is above the graph of 2x. Shade that region to find the solution.

That's it! If you have any more math questions, feel free to ask.