Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Hello, Can you review my answer for this problem. I’m not sure of my work especially A and D Thank you
**Part Three** Find the inverse of each function below. Check your answers in Desmos. a. \( f(x) = 2x^2 - 4 \) - \( y = 2x^2 - 4 \) - \( x = 2y^2 - 4 \) - \( x + 4 = 2y^2 \) - \( \frac{x + 4}{2} = y^2 \) - Solving for \( y \): \( y = \sqrt{\frac{x + 4}{2}} \) b. \( f(x) = \frac{3}{x} + 2 \) - \( y = \frac{3}{x} + 2 \) - \( y - 2 = \frac{3}{x} \) - \( \frac{3}{y - 2} = x \) c. \( f(x) = \frac{2x - 9}{x + 2} \) - \( y = \frac{2x - 9}{x + 2} \) - Cross-multiply: \( x(y + 2) = 2y - 9 \) - \( xy + 2x = 2y - 9 \) - Rearrange terms: \( x(y + 2) = 2y + 9 \) - Solve for \( x \): \( x = \frac{2y + 9}{y + 2} \) d. \( f(x) = a(x - h)^2 + k \) - \( y = a(x - h)^2 + k \) - Solve: \( y - k = a(x - h)^2 \) - \( \frac{y - k}{a} = (x - h)^2 \) - \( \sqrt{\frac{y - k}{a}} + h = x \) - So the inverse: \( f^{-1}(x) = \sqrt{\frac{x - k}{a}} + h \) **Part Four** Graph \( y = \log_3 x \) and \( y = 3^x \) on the same axis using a table. You may want to use two colors to separate the graphs. a) Table and Graph: - For \( y = \log_3 x \) - \( f(1) = \log_
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Transcribed Image Text:**Part Three** Find the inverse of each function below. Check your answers in Desmos. a. \( f(x) = 2x^2 - 4 \) - \( y = 2x^2 - 4 \) - \( x = 2y^2 - 4 \) - \( x + 4 = 2y^2 \) - \( \frac{x + 4}{2} = y^2 \) - Solving for \( y \): \( y = \sqrt{\frac{x + 4}{2}} \) b. \( f(x) = \frac{3}{x} + 2 \) - \( y = \frac{3}{x} + 2 \) - \( y - 2 = \frac{3}{x} \) - \( \frac{3}{y - 2} = x \) c. \( f(x) = \frac{2x - 9}{x + 2} \) - \( y = \frac{2x - 9}{x + 2} \) - Cross-multiply: \( x(y + 2) = 2y - 9 \) - \( xy + 2x = 2y - 9 \) - Rearrange terms: \( x(y + 2) = 2y + 9 \) - Solve for \( x \): \( x = \frac{2y + 9}{y + 2} \) d. \( f(x) = a(x - h)^2 + k \) - \( y = a(x - h)^2 + k \) - Solve: \( y - k = a(x - h)^2 \) - \( \frac{y - k}{a} = (x - h)^2 \) - \( \sqrt{\frac{y - k}{a}} + h = x \) - So the inverse: \( f^{-1}(x) = \sqrt{\frac{x - k}{a}} + h \) **Part Four** Graph \( y = \log_3 x \) and \( y = 3^x \) on the same axis using a table. You may want to use two colors to separate the graphs. a) Table and Graph: - For \( y = \log_3 x \) - \( f(1) = \log_
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