{"id":38831,"date":"2025-06-26T13:27:29","date_gmt":"2025-06-26T13:27:29","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=38831"},"modified":"2025-06-26T13:27:30","modified_gmt":"2025-06-26T13:27:30","slug":"no-calculator-is-allowed-for-this-question-2","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/no-calculator-is-allowed-for-this-question-2\/","title":{"rendered":"NO CALCULATOR IS ALLOWED FOR THIS QUESTION"},"content":{"rendered":"\n
Progress Check: FRQ Part B Name NO CALCULATOR IS ALLOWED FOR THIS QUESTION: Show the question may not explicitly remind you to do so. Clearly show all of your work, even though justifications require that you label any functions, graphs, tables, or other objects that you use. Give mathematical reasons and verify the needed conditions under which relevant theorems, properties, definitions, or tests are applied. Answers without correctness and completeness of your methods as well as your supporting work will usually not receive credit. Algebraic expressions need not be simplified. If not otherwise specified, answers (numeric) after the answer is given as a decimal approximation; it should be correct to three decimal places. The function f is assumed to be the set of all real numbers for which f(r) is a real number. Consider the curve given by the equation (2y + 1)^2 = -3. Show that \u00e2\u02c6\u0161(2y + 1) = \u00c2\u00b1\u00e2\u02c6\u0161(-3). Please respond on separate paper, following directions from your teacher. Write an equation for the line tangent to the curve at the point (-1, 2). Please respond on separate paper, following directions from your teacher. Evaluate at the point (-1, -2) d^2. Please respond on separate paper, following directions from your teacher. The point (4, 0) is on the curve. Find the value of (y – 1)'(0).<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Given:<\/h3>\n\n\n\n
The curve is defined by the equation:(2y+1)2=\u22123(2y + 1)^2 = -3(2y+1)2=\u22123<\/p>\n\n\n\n
\n\n\n\n
Part 1: Show that 2y+1=\u00b1\u22123\\sqrt{2y + 1} = \\pm \\sqrt{-3}2y+1\u200b=\u00b1\u22123\u200b<\/h3>\n\n\n\n
We begin by taking the square root of both sides of the given equation:(2y+1)2=\u22123(2y + 1)^2 = -3(2y+1)2=\u22123<\/p>\n\n\n\n
Take the square root of both sides:(2y+1)2=\u22123\\sqrt{(2y + 1)^2} = \\sqrt{-3}(2y+1)2\u200b=\u22123\u200b<\/p>\n\n\n\n
This gives:\u22232y+1\u2223=\u22123|2y + 1| = \\sqrt{-3}\u22232y+1\u2223=\u22123\u200b<\/p>\n\n\n\n
Since \u22123\\sqrt{-3}\u22123\u200b is not a real number, we write:2y+1=\u00b1\u22123=\u00b1i32y + 1 = \\pm \\sqrt{-3} = \\pm i\\sqrt{3}2y+1=\u00b1\u22123\u200b=\u00b1i3\u200b<\/p>\n\n\n\n
So, the square root expression 2y+1\\sqrt{2y + 1}2y+1\u200b equals \u00b1\u22123\\pm \\sqrt{-3}\u00b1\u22123\u200b, which confirms the answer.<\/p>\n\n\n\n
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Part 2: Write an equation for the line tangent to the curve at the point (-1, 2)<\/h3>\n\n\n\n
Let\u2019s find the derivative using implicit differentiation of the original equation:(2y+1)2=\u22123(2y + 1)^2 = -3(2y+1)2=\u22123<\/p>\n\n\n\n
Differentiate both sides implicitly with respect to xxx:<\/p>\n\n\n\n
Left-hand side using chain rule:2(2y+1)(2\u22c5dydx)=02(2y + 1)(2 \\cdot \\frac{dy}{dx}) = 02(2y+1)(2\u22c5dxdy\u200b)=0<\/p>\n\n\n\n
Now solve for dydx\\frac{dy}{dx}dxdy\u200b:dydx=0(as long as 2y+1\u22600)\\frac{dy}{dx} = 0 \\quad \\text{(as long as } 2y + 1 \\ne 0\\text{)}dxdy\u200b=0(as long as 2y+1\ue020=0)<\/p>\n\n\n\n
At the point (\u22121,2)(-1, 2)(\u22121,2), plug in y=2y = 2y=2:2(2)+1=5\u22600\u21d2dydx=02(2) + 1 = 5 \\ne 0 \\Rightarrow \\frac{dy}{dx} = 02(2)+1=5\ue020=0\u21d2dxdy\u200b=0<\/p>\n\n\n\n
So, the slope of the tangent line is 0. The equation of the tangent line is:y\u22122=0(x+1)\u21d2y=2y – 2 = 0(x + 1) \\Rightarrow y = 2y\u22122=0(x+1)\u21d2y=2<\/p>\n\n\n\n
\n\n\n\n
Part 3: Evaluate at the point (-1, -2) d2ydx2\\frac{d^2y}{dx^2}dx2d2y\u200b<\/h3>\n\n\n\n
Start again from:(2y+1)2=\u22123\u21d2This equation is not valid for real numbers when y=\u22122\u21d22(\u22122)+1=\u22124+1=\u22123\u21d2(\u22123)2=9\u2260\u22123(2y + 1)^2 = -3 \\Rightarrow \\text{This equation is not valid for real numbers when } y = -2 \\Rightarrow 2(-2) + 1 = -4 + 1 = -3 \\Rightarrow (-3)^2 = 9 \\ne -3(2y+1)2=\u22123\u21d2This equation is not valid for real numbers when y=\u22122\u21d22(\u22122)+1=\u22124+1=\u22123\u21d2(\u22123)2=9\ue020=\u22123<\/p>\n\n\n\n
So, the point (\u22121,\u22122)(-1, -2)(\u22121,\u22122) is not on the curve. Therefore, d2ydx2\\frac{d^2y}{dx^2}dx2d2y\u200b is not defined at that point.<\/p>\n\n\n\n
\n\n\n\n
Part 4: The point (4, 0) is on the curve. Find the value of (y\u22121)\u2032(0)(y – 1)'(0)(y\u22121)\u2032(0)<\/h3>\n\n\n\n
Here, we need to find the derivative of y\u22121y – 1y\u22121 with respect to xxx, then evaluate at x=0x = 0x=0. Since:(y\u22121)\u2032=dydx(y – 1)’ = \\frac{dy}{dx}(y\u22121)\u2032=dxdy\u200b<\/p>\n\n\n\n
We already found that:dydx=04(2y+1)=0 (from earlier)\\frac{dy}{dx} = \\frac{0}{4(2y + 1)} = 0 \\text{ (from earlier)}dxdy\u200b=4(2y+1)0\u200b=0 (from earlier)<\/p>\n\n\n\n
At point (4, 0), we plug y=0y = 0y=0 into the derivative expression:<\/p>\n\n\n\n
This problem involves implicit differentiation and analysis of a non-standard curve. First, the equation (2y+1)2=\u22123(2y + 1)^2 = -3(2y+1)2=\u22123 tells us the square of a real quantity is equal to a negative number, which is not possible in the real number system. Taking the square root of both sides, we find 2y+1=\u00b1\u22123\\sqrt{2y + 1} = \\pm \\sqrt{-3}2y+1\u200b=\u00b1\u22123\u200b, which is an imaginary number. This suggests the curve defined does not intersect the real plane unless imaginary values are considered.<\/p>\n\n\n\n
When asked to find the tangent line at the point (\u22121,2)(-1, 2)(\u22121,2), we must verify that this point satisfies the equation. Substituting y=2y = 2y=2 gives (2\u22c52+1)2=52=25\u2260\u22123(2 \\cdot 2 + 1)^2 = 5^2 = 25 \\ne -3(2\u22c52+1)2=52=25\ue020=\u22123, so this point is not on the curve unless the equation has a typo. However, assuming the point is correct and the implicit differentiation holds, we find that the slope is 0, giving a horizontal tangent line y=2y = 2y=2.<\/p>\n\n\n\n
In part 3, evaluating the second derivative at (\u22121,\u22122)(-1, -2)(\u22121,\u22122) also fails because the point does not lie on the curve. Plugging y=\u22122y = -2y=\u22122 into the curve gives (2\u22c5\u22122+1)2=(\u22123)2=9(2 \\cdot -2 + 1)^2 = (-3)^2 = 9(2\u22c5\u22122+1)2=(\u22123)2=9, which again does not equal \u22123-3\u22123. Therefore, the second derivative is undefined there.<\/p>\n\n\n\n
In part 4, to find the derivative of y\u22121y – 1y\u22121, we simply differentiate to get dydx\\frac{dy}{dx}dxdy\u200b, and evaluate it at the point (4,0)(4, 0)(4,0). This results in 0, since the derivative equation equals zero due to the constant right-hand side.<\/p>\n\n\n\n
Each part relies on implicit differentiation and algebraic substitution, emphasizing the importance of verifying point validity before applying calculus tools.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
Progress Check: FRQ Part B Name NO CALCULATOR IS ALLOWED FOR THIS QUESTION: Show the question may not explicitly remind you to do so. Clearly show all of your work, even though justifications require that you label any functions, graphs, tables, or other objects that you use. Give mathematical reasons and verify the needed conditions […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-38831","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38831","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=38831"}],"version-history":[{"count":2,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38831\/revisions"}],"predecessor-version":[{"id":38838,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38831\/revisions\/38838"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=38831"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=38831"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=38831"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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