# Calculus

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.

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1. (fā¦g)(x) = f(g(x))
f(x) = 10^x, so
f(g) = 10^g = 10^(logx) = x
Similarly, gā¦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

(fā¦g)(-2) = f(g(-2)) = f(6) is undefined
(gā¦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.

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posted by Steve

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