Math 111 A&B Quiz 1 Overview

Math 111 A&B Quiz 1 - Autumn 2025: Detailed Notes

Section 1: Solving Linear Equations

  • Problem 1 (3 points): Solve for x:
    2x−7=42x - 7 = 4

    • Solution Steps:
    1. Add 7 to both sides:
      2x=4+72x = 4 + 7
    2. Simplify:
      2x=112x = 11
    3. Divide both sides by 2:
      x=112x = \frac{11}{2} or x=5.5x = 5.5

  • Problem 2 (6 points): Solve for x:
    2imes3x−1imes5=42 imes 3^x - 1 imes 5 = 4

    • Solution Steps:
    1. Rewrite the equation:
      2imes3x=4+52 imes 3^x = 4 + 5
    2. Combine like terms:
      2imes3x=92 imes 3^x = 9
    3. Divide both sides by 2:
      3x=923^x = \frac{9}{2}
    4. Solve for x using logarithms:
      x=extlog(92)extlog(3)x = \frac{ ext{log}(\frac{9}{2})}{ ext{log}(3)}

  • Problem 3 (3 points): Solve for x:
    2x−4=3x+1−x2x - 4 = 3x + 1 - x

    • Solution Steps:
    1. Combine like terms on the right side:
      2x−4=2x+12x - 4 = 2x + 1
    2. Move 2x from both sides:
      −4=1-4 = 1
    3. This is a contradiction; thus, there is no solution.

  • Problem 4 (10 points): Solve for x:
    3x+2x−1=4\frac{3x+2}{x-1} = 4

    • Solution Steps:
    1. Cross-multiply to eliminate fraction:
      3x+2=4(x−1)3x + 2 = 4(x - 1)
    2. Distribute:
      3x+2=4x−43x + 2 = 4x - 4
    3. Rearrange terms:
      2+4=4x−3x2 + 4 = 4x - 3x
      6=x6 = x
    4. Thus, x=6x = 6.

  • Problem 5: Consider the inequality:
    −3x+2ext≤5-3x + 2 ext{ ≤ } 5

    • (a) (5 points): Solve for x and write your answer as an inequality:
    1. Rearrange the inequality:
      −3xext≤5−2-3x ext{ ≤ } 5 - 2
      −3xext≤3-3x ext{ ≤ } 3
    2. Divide by -3 (flip inequality):
      xext≥−1x ext{ ≥ } -1
    • (b) (3 points): Draw a number line and show your answer on that number line:
      • The number line includes the point -1 shaded towards the right, indicating all values greater than or equal to -1.

Section 2: Evaluating Quadratic Functions

  • Problem 6: Consider the function f(x)=−x2+2x−1f(x) = -x^2 + 2x - 1
    • (a) (3 points): Compute f(1)f(1):
      • Plug in the value:
        f(1)=−12+2(1)−1f(1) = -1^2 + 2(1) - 1
      • Calculation:
        f(1)=−1+2−1=0f(1) = -1 + 2 - 1 = 0
    • (b) (3 points): Compute f(0)f(0):
      • Plug in the value:
        f(0)=−02+2(0)−1f(0) = -0^2 + 2(0) - 1
      • Calculation:
        f(0)=0+0−1=−1f(0) = 0 + 0 - 1 = -1
    • (c) (4 points): Compute f(2−4)f(2 − 4):
      • Simplify first:
        2−4=−22 - 4 = -2
      • Now plug in:
        f(−2)=−(−2)2+2(−2)−1f(-2) = -(-2)^2 + 2(-2) - 1
      • Calculation:
        f(−2)=−4−4−1=−9f(-2) = -4 - 4 - 1 = -9

Section 3: Cost Function of Sprockets

  • Problem 7 (10 points): A company buys sprockets and gets a discount based on the number that they purchase. If x is the number of sprockets purchased, the cost for each sprocket purchased is: extCostpersprocket=200−1.35xext{Cost per sprocket} = 200 - 1.35x
    • Determine the cost function, C(x):
      • The total cost function can be calculated by multiplying the cost per item by the number of items purchased:
      • Thus, the function is:
        C(x)=ximes(200−1.35x)C(x) = x imes (200 - 1.35x)
      • Expanding the equation:
        C(x)=200x−1.35x2C(x) = 200x - 1.35x^2
      • Therefore, the total cost function is: C(x)=200x−1.35x2C(x) = 200x - 1.35x^2.