University:
Carleton College
Course:
MATH 120 | Calculus 2
Academic year:

2024
Views:

329

Pages:

3
Author:

Nevaeh Bauer

Question 1 Write a function based on the given parent function and transformations in the given order. Parent function: y = x³ Shift 4.2 units to the left. Reflect across the y-axis. Shift upward 2.3 units. Answer Shift 4.2 units to the left: Add 4.2 to the input x: y = (x + 4.2)³ Reflect across the y-axis: Change the sign of x: y = (-x + 4.2)³ Shift upward 2.3 units: Add 2.3 to the output: y = (-x + 4.2)³ + 2.3 Question 2 Solve the absolute value equation or indicate the equation has no solution: |2X - 3| = 11 Answer Step 1 Take into account the absolute value equation shown. |2X - 3| = 11 As we know that |x|=c is equivalent to x = c or x = -c where x is an algebraic expression and c is a positive real number. In order to write the given absolute value equation, 2x - 3 = 11 or 2x - 3 = -11 Step 2 now, solve both the equations, 2x - 3 + 3 = 11 + 3 or 2x - 3 + 3 = -11 + 3 2x = 14 or 2x = -8 x = 7 or x = -4 so the solution set is {-4, 7}. Question 3 Solve the absolute value inequality. 8 < |7 - 3x| Answer Step 1 First isolate the absolute value expression to the left side of inequality to solve the inequality. Remove the absolute bars by setting a compound inequality. Then solve the inequalities to obtain the solution. Step 2 The solution to the inequality 8 < |7 - 3x| can be found as follows. First isolate the absolute expression by changing the sides as |7 - 3x| > 8. Now apply absolute rule: if |u| > a, then u < -a or u > a to remove the absolute bars and solve the two inequalities. 7 - 3x < -8 or 7 - 3x > 8 -3x < -15 or -3x > 1 x > 5 or x < -1/3 Hence, the solution to the absolute value inequality 8 < |7 - 3x| is x < -1/3 or x > 5. Question 4 Find the absolute maximum value and the absolute minimum value, if any, of the function. g(x) = -x² + 2x + 6 Answer Step 1 Given g(x) = -x² + 2x + 6 To find out absolute max or min, let find out derivative g'(x) g'(x) = -2x + 2 Now set the derivative = 0 and solve for x -2x + 2 = 0 -2x = -2 x = 1 Step 2 To check whether x = 1 is maximum or minimum we use second derivative test g'(x) = -2x + 2, g''(x) = -2 second derivative is negative so f(x) is maximum at x = 1 There is no absolute minimum Now find out absolute maximum value at x = 1 Substitute x = 1 and find out g(1) g(1) = -(1)² + 2(1) + 6 g(1) = 7 So absolute maximum value is 7 Absolute minimum does not exists Question 5 Find the linear approximation of the function f(x) = √(1 - x) at a = 0 and use it to approximate the numbers √0.9 and √0.99. Answer The formula for linearization L(x) is L(x) = f(a) + f'(a)(x – a) we have f(x) = √(1 - x) with a = 0 f(a) = √1 - 0 = 1 f'(x) = (√(1 - x))' = (1 - x)¹/² = (1/2)(1 - x)⁻¹/² (-1) = -1/2√(1-x) f'(a) = -1/2√1 - 0 = -1/2 L(x) = 1 - 1/2(x - 0) = 1 - 1/2 x √1 - x ≈ √0.99 so x = 0.01 L(0.01) = 1 - 1/2 * 0.01 = 0.995 √1 - x ≈ √0.999 so x = 0.001 L(0.01) = 1 - 1/2 * 0.001 = 0.9995