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Test 2 ![]() Solution (1.9)2 + 8(1.9) – 2 = 16.810 (1.99)2 + 8(1.99) – 2 = 17.880 and so on Complete the table and use the result to find the indicated limit. ![]() Solution Enter in ti-nspire or ti84 plus exactly like so (.94 – 1) ÷ (.9 – 1) = 3.439 (.99 - 1) ÷ .99 – 1) = 3.940 And so on… Use the properties of limits to help decide whether the limit exists. If the limit exists, find its value. ![]() Solution Simplify x2 – 9 / x – 3 = x / x2 = x and -9 / -3 = 3 Now put the equation back together = x + 3 Now plug in the value of x --> 3 (3) + 3 = 6 Use the properties of limits to help decide whether the limit exists. If the limit exists, find its value. ![]() The Ti-nspire will solve this problem if you are allowed to use it. Otherwise, solve like so Solution Simplify ![]() x + 6 + 5 ![]() x + 1 Now plug in the value of (x à-5) 1 – 5 = -4 Use the properties of limits to help decide whether the limit exists. If the limit exists, find its value. ![]() ![]() Solution divide by the highest denominator power ![]() ![]() ![]() ![]() ![]() Use the properties of limits to help decide whether the limit exists. If the limit exists, find its value. ![]() Solution divide by the highest denominator power ![]() ![]() ![]() ![]() ![]() ![]() ![]() The average cost per tape, denoted by c (x) is found by dividing c(x) by x. Find lim c(x) x--> 2.000 ![]() The equation we need to write is ![]() ![]() Then we plug in all the values of x = 2,000 10,000 + 7(2,000) ÷ 2,000 10,000 + 14,000 ÷ 2,000 24,000 ÷ 2,000 = 12 Find all values x = a where the function is discontinuous. ![]() ![]() Find the average rate of change for the function over the given interval. y = x2 + 4x between x = 4 and x = 8. ![]() Solve the long way… ![]() 64 + 32 – 16 + -16 = 96 – 32 = 64 64 / 8 – 4 = 64 / 4 = 16 Or This is a cool online calculator that will solve this problem for you. https://www.emathhelp.net/calculators/calculus-1/average-rate-of-change-calculator/ enter equation and points. ![]() Find f’(x) at the given value of x. f(x) x2 – 9x + 2, find f’(- 4). ![]() Solve the long way… nn-1 (n = the exponent) If a variable has no exponent, then the variable is cancelled out. For example, with 2x, the derivative would = 2 because the exponent is equal to 1 and n-1 would cancel it out. If a number in an equation has no variable, then the derivative = 0. What I do is I always do is multiply a number with no variable by 0 when finding the derivative. For example, 1, the derivative = 0 (you do not need to represent the 0) So, the derivative of x2 – 9x + 2 is: 2(x2 – 1) – 1(91 – 1) + 2(0) 2x – 9 2(-4) – 9 = -8 – 9 = -17 Or This calculator will find the derivative for you. https://www.derivative-calculator.net/ ![]() Using the definition of the derivative, find f’(x). Then find f’(1), f’(2) and f’(3) when the derivative exists. f(x) = -x2 + 9x - 4 ![]() ![]() Solve Find the derivative of f(x) = -x2 + 9x – 4 (which is -2x + 9) and plug in the values of f’(1) which is 1, f’(2) which is 2 and f’(3) which is 3. This is the same as above, except you need to find the derivative and find multiple values for f’(). solve the long way. nn-1 (n = the exponent) The same rules apply as above. nn-1(-x2 + 9x – 4) = 2(x2 – 1) + 9(x1 – 1) – 4(0) = -2x + 9 Now plug in all the values of f’() which are 1, 2, & 3. -2(1) + 9 = -2 + 9 = 7 -2(2) + 9 = -4 + 9 = 5 -2(3) + 9 = -6 + 9 = 3 This calculator will find the derivative for you. https://www.derivative-calculator.net/ and enter in the equation as so ![]() the price of the item. Find the rate of change of demand with respect to price. ![]() We need to find the demand which is simply the derivative in this case. The book gave long lessons on finding the demand however after I solved about a dozen of the problems, I came to realize that the demand was simply the derivative of the expression and the variable used is p instead of x. Imagine the time I saved after that! Find the derivative of -4p2 + 3p + 7 = -8p + 3 The same rule applies as above. nn – 1(-4p2 + 3p +7) = 2(-4p2 – 1) + 3(p1 – 1) + 7(0) D’(p)-8p + 3 It is as simple as that! Homework Chapter 1A 1B 2A 2B 3A 3B 4A 4B 5A 6A Chapter Tests 1 2 3 4 5 6 7 FINAL EXAM | Need A Tutor? | Need Homework Help?
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