Math 19: Calculus Winter 2008 Instructor: Jennifer KlokeSolutions to the Second Set of Practice Problems1. Write as a single fraction:3x(x − 3)−2x(x − 3)(x + 2)+1x.Solution.3x(x − 3)−2x(x − 3)(x + 2)+1x=3(x + 2)x(x − 3)(x + 2)−2x(x)x(x − 3)(x + 2)+(x + 2)(x − 3)x(x + 2)(x − 3)=3(x + 2) − 2x2+ (x + 2)(x − 3)x(x − 3)(x + 2)=3x + 6 − 2x2+ x2− x − 6x(x − 3)(x + 2)=−x2+ 2xx(x − 3)(x + 2)=x(−x + 2)x(x − 3)(x + 2)=−x + 2(x − 3)(x + 2)2. Simplify8x3+xexx tan(x)√4x.Solution.8x3+xexx tan(x)√4x=8x3+ xex1x tan(x)√4x=x(8x2+ 1)exx tan(x)(2√x)=8x2+ 1extan(x)(2√x)3. Find all the solutions to |x2− 3| = 1.1Solution.|x2− 3| = 1x2− 3 = ±1x2= 4, 2The solutions to the above equality are the solutions to x2= 4 and the solutions to x2= 2.The solutions to x2= 4 are x = 2 and x = −2 . The solutions to x2= 2 are x =√2 andx = −√2 .4. Find all the solutions to (x2− 4)(x2+ 5x + 4) = 0.Solution. The solutions to (x2− 4)(x2+ 5x + 4) = 0 are the solutions to x2− 4 = 0 and thesolutions to x2+ 5x + 4 = 0. The solutions of x2−4 = 0 are x = 2 and x = −2 . Similarly,the solutions of x2+ 5x + 4 = 0 are x = −1 and x = −4 (use the quadratic formula if youdon’t see how to factor it otherwise).5. Find the equation of the line that passes through the points (π, 0) and (0,5).Solution. y − 0 = (5−00−π)(x − π)y = (−5/π)(x − π)6. Find the equation of the line that passes through the point (5,-1) and has slope e.Solution. y − (−1) = e(x − 5)y + 1 = e(x − 5)7. Find the y-intercepts and x-intercepts of f(x) = ex+2− 2.Solution. The y-intercept is f(0) = e0+2−2 = e2− 2 . The x-intercepts are the solutions tothe equation 0 = ex+2− 2.ex+2− 2 = 0ex+2= 2ln(ex+2) = ln(2)x + 2 = ln(2)x = ln(2) − 28. Find the y-intercepts and x-intercepts of f(x) = (3 − x)(ln(x) + 1).2Solution. The y-intercept is f(0) = (3 − 0)(ln(0) + 1) = (3)(1 + 1) = 6 . The x-interceptsare the solutions to the equation 0 = (3 − x)(ln(x) + 1). The expression 3 − x equals 0when x = 3 . The expression ln(x) + 1 equals 0 when ln(x) = −1 which happens whenx = eln(x)= e−1.9. If tan θ = 3/4, find sin θ.Solution. Draw a right triangle with an angle θ, the side opposite to θ with length 3, andthe side adjacent to θ with length 4. (Recall that tan is “opposite over adjacent”.) Thehypotenuse must then have length√42+ 32=√25 = 5. Thus sin θ =35.10. What is the value of sin(5π/6), cos(7π/4), csc(π/2) and cot(π/6)?Solution. Using the unit circle like we did in class on Friday, we can determine the followingvalues:sin(5π6) =12cos(7π4) =√22csc(π2) =1sin(π/2)= 1/1= 1cot(π6) =cos(π/6)sin(π/6)=√3/21/2=√311. Find all the solutions to ln((2x2− 1)) = 0.Solution.ln((2x2− 1)) = 0eln((2x2−1))= e02x2− 1 = 12x2= 2x2= 1x = ±1 .312. Simplify ln(e2x√xx−√x).Solution.ln(e2x√xx −√x) = ln(e2x√x) − ln(x −√x)= ln(e2x) + ln(√x) − ln(x −√x)= 2x + (1/2) ln(x) − ln(x −√x) .13. Write the domain of y =√x +1x−2in interval notation.Solution. We know that the domain of√x is all x ≥ 0. We also know that the domain of1x−2is all x except x = 2. By combining these two facts, we find that the domain of y =√x +1x−2is all x ≥ 0 except x = 2. In interval notation, the domain is [0, 2) ∪ (2, ∞) .14. Let f(x) = tan x and g(x) = 8x2+ e1. Find (g ◦ f)(x).Solution.(g ◦ f)(x) = g(f(x))= g(tan(x))= 8(tan(x))2+ e= 8 tan2(x) + e15. Write 2ln(4x)as a composition of simpler functions.Solution. Let f(x) = 2x, g(x) = ln(x), and h(x) = 4x.2ln(4x)= f(ln(4x))= f(g(4x))= f(g(h(x)))= (f ◦ (g ◦ h))(x)= ((2x) ◦ ((ln(x)) ◦ (4x)))(x)16. Graph f(x) =ex: x ≤ 01 : 0 < x < 1−x2: x ≥ 1.4Solution. See the graphs linked from the website.17. Draw the graph of y = ln x.Solution. See the graphs linked from the website.18. Draw the graph of y = x2− 4x + 3.Solution. See the graphs linked from the