Math 1300 - 401 1.1 - 1.4 Practice Problems Name: 1. Consider the ...

Math 1300 - 401 1.1 - 1.4 Practice Problems Name:1. Consider the functionf(x) = ln(x)(a) What is the domain of this function?(0, ∞)(b) What is the range?R (all real numbers)(c) Is this function invertible? If so, what is its inverse? (Justify your answer.)Yes, it is invertible since it is a one-to-one function and passes the horizontal line test.The inverse is g(x) = e x . To justify this, we’ll compose f and g in both directions andshow that we get x either way:f(g(x)) = f(e x ) = ln(e x ) = x ln(e) = x · 1 = xg(f(x)) = g(ln(x)) = e ln(x) = x2. Suppose we define a function f that take names of countries as inputs and gives the country’sofficial language as an output. For instance,f(Brazil) = PortugueseExplain why this function is not invertible.This function is not invertible since it is not one-to-one. For instance, the official languageof the USA is English, and the official language of England is also English. So an inversefunction wouldn’t know what country should be an output for the input “English”.3. The graph of the function y = x 2 is shifted up by 2 and right by 5. Write an expression forthe new function.y = (x − 5) 2 + 24. Draw and label a set of (x, y)-axes in the space below. On them, plot the line given by theequation6x + 2y = 4What is the slope of this line? What is its y-intercept?We solve for y to gety = −3x + 2so this is a line with slope -3 and y-intercept 2.

Math 1300 - 401 1.1 - 1.4 Practice Problems Name:5. The population of Logopolis, USA is 20 mathematicians at time t = 0. Write a formula thatmodels the population of Logopolis as a function of time if . . .(a) . . . the population is increasing at a rate of 10 mathematicians per year.P = 20 + 10t (where t is in years)(b) . . . the population is increasing at a rate of 10% per year.P = 20(1.1) t (where t is in years)(c) In case (b), how long will it take for the population of Logopolis to double?After the doubling time passes, the population will be 40 mathematicians. So we set Pequal to 40 and solve for t:40 = 20(1.1) t2 = (1.1) tlog(2) = log((1.1) t )log(2) = t log(1.1)t = log(2)/ log(1.1)t ≈ 7.3 years6. Solve for x:5(3) 7x = 8(3) 5x5(3) 7x = 8(3) 5x(3) 7x(3) 5x = 8 5(3) 7x−5x = 8/5(3) 2x = 8/5log((3) 2x ) = log(8/5)2x log(3) = log(8/5)2x = log(8/5)log(3)x = log(8/5)2 log(3)x ≈ 0.214

Math 1300 - 401 1.1 - 1.4 Practice Problems Name:For the problems below, you don’t need to simplify.7. If f(x) = 3x 2 + x + 5 and g(x) = e x , thenf(g(x)) = 3(e x ) 2 + e x + 5andg(f(x)) = e (3x2 +x+5)8. If h(x) = √ x + x and j(x) = x − 7, thenh(j(x)) = √ x − 7 + (x − 7)andj(h(x)) = ( √ x + x) − 7