D f = (2, ) (x + 1)(x 3) (b) g(x) = x 1 solution: We need the thing inside the root to be greater than or equal to 0. So we set up a sign table.

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1 . Find the domains of the following functions: (a) f(x) = ln(x ) We need x > 0, or x >. Thus D f = (, ) (x + )(x 3) (b) g(x) = x We need the thing inside the root to be greater than or equal to 0. So we set up a sign table. x (, ) (, ) (, 3) (3, ) x x + + x 3 + (x + )(x 3) x + + Thus the expression inside the root is greater than or equal to 0 on [, ) [3, ). Thus D g = [, ) [3, ) (c) h(x) = sin(x) cos(x) Here we need to exclude all the points where cos or sin are 0. sin is 0 at all multiples of π and cos is 0 at π plus all multiples of π. Together this means we need to exclude all integer multiples of π: D h = {x R x nπ }, n Z x (d) k(x) = x x + If we calculate the discriminant of the quadratic under the root: = b 4ac = ( ) 4()() = 7 Which means that the polynomial has no roots. Since it opens up, that means it is always positive. Hence its root is always defined, and its root is never 0, and so the domain is every real number: D k = R (e) r(x) = sin ( ) π x The only problem here is that we can t divide by 0. Hence only x = 0 is excluded: D r = {x R x 0}

2 . Let f(x) = 3 3x + 9 and g(x) = x + 6. Find f g(x), g f(x), g (x), and f (x). f g(x) = f(g(x)) = f(x + 6) = 3 3(x + 6) + 9 = 3 6x + 7 g f(x) = g(f(x)) = g( 3 3x + 9) = 3 3x To find the inverse functions we switch x and y and then solve for y. Thus f (x) = x Thus g (x) = x 6. x = 3 3y + 9 x 3 = 3y + 9 y = x3 9 3 x = y + 6 y = x 6 y = x 6

3 3. Solve the following: (a) 5 ln(x + 3) = 0 for x 5 ln(x + 3) = 0 5 ln(x + 3) = ln(x + 3) = 5 x + 3 = e /5 (b) log (y) = log (y ) log (y) + for y. x = e /5 3 log (y) log (y ) + log (y) = ( ) y y log = y y y y = log (y) = log (y ) log (y) + y = 4y 4 y = 4 y = ± Thus we have two solutions, y = ±. However, only y = + is in the domain of the original expression, for if you plug a negative number into log y you get something that is not defined. Thus our only solution is y =. (c) 3 t+3 = 5 t for t. 3 t+3 = 5 t ln(3 t+3 ) = ln(5 t ) Of course any log works here. (t + 3) ln 3 = t ln 5 t ln 5 t ln 3 = 3 ln 3 t(ln 5 ln 3) = 3 ln 3 t = 3 ln 3 ln 5 ln 3

4 4. Find all solutions of sin(3x) = in the interval [0, π]. An angle in the first quadrant that has sin equal to is π. sin is also positive 6 in the second quadrant, meaning our second basic solution is 5π. Thus 6 3x = π 6 + kπ, k Z or 3x = 5π 6 + kπ, k Z x = π 8 + kπ 5π, k Z or x = kπ 3, k Z We solve the following inequalities to find the values of k we need 0 π 8 + kπ 3 π k 3 8 k 3 8 = k 6 6 k 7.4 The ks in this range are k = 0 and k =. Hence For the other solutions we have x = π 8, 3π 8 0 5π 8 + kπ 3 π k k = k k 3.08 The ks in this range are k = 0 and k =. Hence x = 5π 8, 7π 8

5 5. Sketch the following functions: (a) f(x) = x + x This is a rational function. It has no x-intercepts because the top is never 0. Substituting in 0 for x shows us that the y-intercept is. The horizonal asymptote is at the ratio of the leading coefficients, i.e. at y = =. Since the roots of the denominator are ±, there are vertical asymptotes at x = ±. Looking at a sign table: x (, ) (, ) (, ) x x + x x + x + + Since f is positive to the left of the asymptote at, it has to approach as it approaches from the left. Since f is negative to the right of the asymptote at, it has to approach as it approaches from the right. Since f is negative to the left of the asymptote at, it has to approach as it approaches from the left. Finally, since f is positive to the right of the asymptote at, it has to approach as it approaches from the right. This all gets put together to give (b) g(x) = 3x+5 x+

6 Using long division, 3 x + ) 3x + 5 3x 3 Thus g(x) = 3x + 5 x + = 3 + x +. Hence the graph of g looks like the graph of x shifted up by 3, to the left by, and stretched vertically by. (c) h(x) = 3 cos (x) This is a usual cos graph stretched vertically by 3. (d) k(x) = 3 sin ( ( x π 3 )) + This is a sin graph with amplitude 3 and period π shifted up by and to the right by π 3.

7 (e) r(x) = tan(x) This is a tan graph flipped vertically and shifted up by units.

8 6. Verify the following identities: (a) cos(sin (x)) = x (b) tan θ = sec θ cos θ csc θ LHS = cos(sin (x)) = sin (sin (x)) = x = RHS (c) tan x tan y = cos(x + y) cos x cos y RHS = sec θ cos θ = cos θ cos θ = cos θ cos θ = sin θ cos θ = sin θ cos θ sin θ = sin θ cos θ csc θ = tan θ csc θ = LHS RHS = cos(x + y) cos x cos y = cos x cos y sin x sin y cos x cos y sin x sin y = cos x cos y ( ) ( ) sin x sin y = cos x cos y = tan x tan y = LHS

9 7. If r = 3% and P = $0, 000, how much do you have after 0 years if interest is compounded monthly? Continuously? If interest is compounded continuously, how long does it take you to reach $35,000? The equation for discrete compound interest is ] nt [ A(t) = P + r n [ A(t) = [ A(0) = $3494 The equation for continuous compound interest is ] t ] (0) A(t) = P e rt A(t) = 0000e 0.03t A(0) = 0000e 0.3 $3498 If interest is compounded continuously to a target of $35,000, we have A(t) = 0000e 0.03t = 0000e 0.03t 3.5 = e 0.03t 0.03t = ln 3.5 t = ln y

10 8. Find the exact values of (a) tan 5π 5π = π 6 + π 4, so tan 5π = tan( π 6 + π 4 ) = tan π 6 + tan π 4 tan π 6 tan π 4 = / 3 + / 3 = ( + 3)/ 3 ( 3 )/ 3 = ( + 3) ( 3 ) (b) cos 5π + cos π Using the sum-to-product formula, (c) csc 0.5 ( ) ( ) x + y x y cos x + cos y = cos cos cos 5π + cos π ( 5π = cos + π ) ( 5π cos π ) ( π ) ( π ) = cos cos 4 6 ( ) ( ) 3 = 6 = 0.5 lies in the third quadrant and 0.5 = Thus, since

11 csc is negative in quadrant III, csc 0.5 = csc.5 csc 0.5 = csc.5 = cos.5 = cos(45 /) = +cos 45 = + / = ( + )/ 4 = + 9. If revenue is given by R(x) = 80x 0.4x, where x is the number of units sold, what is the maximum revenue and how many units do you have to sell to get it? (Hint: it is a quadratic) We put this in standard form 80x 0.4x = 0.4(x 00x) = 0.4(x 00x ) = 0.4(x 00x ) = 0.4(x 00) This is in standard form. The maximum it attains is 4000 at the value x = 00. Thus the maximum revenue is $4000 and it as attained when you sell 00 units.

12 0. (a) If f(x) = x +, use the definition of the derivative to find f (x). We go straight to the definition f f(x + h) f(x) (x) = lim h 0 h x + h + x + = lim h 0 h ( ) x + h + x + x + h + + x + = lim h 0 h x + h + + x + x + h + (x + ) = lim h 0 h( x + h + + x + ) h = lim h 0 h( x + h + + x + ) = lim h 0 x + h + + x + = x + + x + = x + (b) Find the following limits. lim x x x Both the top and bottom are 0 at the limit point, so there is some hope. Try to rationalize the numerator: x lim x x ( ) x x + = lim x x x + x = lim x (x )( x + ) = lim x x + = x lim 8 x x Here the bottom is 0 at the limit point but the top is not. That means that x = is a vertical asymptote for this function and hence the limit does not exist.