University:
Stanford University
Course:
MATH 19 | Calculus
Academic year:

2023
Views:

14

Pages:

3
Author:

MathPaladin

Questions and Answers Sheet 1 Inequalities Systems and Graphs Question #1 Graph equation. 2𝑥 − 3𝑦 = 6 Answer: We can graph an equation by determining the x- and y- intercepts and then drawing the line through them. Set 𝑥 = 0 to find the y-intercept 2 ⋅ 0 − 3𝑦 = 6 0 − 3𝑦 = 6 𝑦 = −2 Set 𝑦 = 0 to find the x-intercept 2𝑥 − 3 ⋅ 0 = 6 2𝑥 − 0 = 6 𝑥=3 Question #2 Find the slope and y-intercept of the line. Graph the line 5𝑥 + 𝑦 = 5 Answer: The slope-intercept form is 𝑦 = 𝑚𝑥 + 𝑏, where m is the slope and b is the y-intercept. Given straight line −5𝑥 + 𝑦 = 5 can be written as, 𝑦 = 5𝑥 + 5 Now comparing above equation with slope-intercept form 𝑦 = 𝑚𝑥 + 𝑐 We get, slope = 𝑚 = 5 and y-intercept = 𝑐 = 5. Question #3 Suggest a coordinate plane located at 𝑊(−3,5) and 𝑍(1,1) with 𝑊𝑍. 𝑊𝑍 becomes overline{W'Z'} , where 𝑊 ′ (−12,20) and 𝑍 ′ (4,4), under a dilation centered at the origin. What is this dilation's scale factor? Answer: If the (x,y) point is dilated by a scale factor of k centered at the origin, then the image has (𝑘𝑥, 𝑘𝑦) coordinates. Given: 𝑊(−3,5) 𝑊 ′ (−12,20) Since -12 and 20 are both products of -3 and 5 multiplied by 4 the, then the scale factor has to be 𝑘 = 4. To verify this, check the other points: 𝑍(1,1) 𝑍 ′ (4,4) Since 4 is 1 multiplied by 4, it's verified that the scale factor is 𝑘 = 4. Question #4 Solve (𝑥 + 1)(𝑥 − 3) > 0. Answer: Zeros are: 𝑥 = −1 , 𝑥 = 3 𝑥 < −1 𝑥>3 Question #5 Use the given graph off to find a number 𝛿 such that if |𝑥 − 1| < 𝛿 then |𝑓(𝑥) − 1| < 0.2 Answer: |𝑋| < 𝑎 implies −𝑎 < 𝑋 < 𝑎 Add 1 in each and simplify |𝑓(𝑥) − 1| < 0.2 −0.2 < 𝑓(𝑥) − 1 < 0.2 0.8 < 𝑓(𝑥) < 1.2 The graph shows that if 0.8 < 𝑓(𝑥) < 1.2 x must verify 0.7 < 𝑥 < 1.1 On the other hand we have x that must verify this unequality: |𝑥 − 1| < 𝛿 −𝛿 < 𝑥 − 1 < 𝛿 If we can find \delta that verify −0.3 ≤ −𝛿 < 𝑥 − 1 < 𝛿 ≤ 0.1 , we can make sure that 0.7 < 𝑥 < 1.1 so 𝛿 must verify two conditions: 𝛿 ≤ 0.1 and −0.3 ≤ −𝛿 which means that 𝛿 ≤ 0.1 and 0.3 ≥ 𝛿 0.7 < 𝑥 < 1.1 0.7 − 1 < 𝑥 − 1 < 1.1 − 1 −0.3 < 𝑥 − 1 < 0.1 We can choose 𝛿 = 0.1 , so to be sure two conditions are satisfied, but if you want, you can choose any value.