Question

In Problems 9-52, determine where the function is (a) increasing; (b) decreasing; and (c) determine where relative extrema occur. Do not sketch the graph. 16. $y = -3 + 12x - x^3$ 18. $y = x^3 - 6x^2 + 12x - 6$ 30. $y = \sqrt[3]{x(x - 2)}$ 52. $y = (x^2 + 1)e^{-x}$ 38. $y = \frac{2x^2}{4x^2 - 25}$ 40. $y = \sqrt[3]{x^3 - 9x}$ 42. $y = x^2(x + 3)^4$ 44. $y = (1 - x)^{2/3}$ 46. $y = x \ln x$ 48. $y = x^{-1}e^x$

          In Problems 9-52, determine where the function is (a) increasing;
(b) decreasing; and (c) determine where relative extrema occur.
Do not sketch the graph.
16. $y = -3 + 12x - x^3$
18. $y = x^3 - 6x^2 + 12x - 6$
30. $y = \sqrt[3]{x(x - 2)}$
52. $y = (x^2 + 1)e^{-x}$
38. $y = \frac{2x^2}{4x^2 - 25}$
40. $y = \sqrt[3]{x^3 - 9x}$
42. $y = x^2(x + 3)^4$
44. $y = (1 - x)^{2/3}$
46. $y = x \ln x$
48. $y = x^{-1}e^x$

In Problems 9-52, determine where the function is (a) increasing;
(b) decreasing; and (c) determine where relative extrema occur.
Do not sketch the graph.
16. y = -3 + 12x - x^3
18. y = x^3 - 6x^2 + 12x - 6
30. y = √(x(x - 2))
52. y = (x^2 + 1)e^-x
38. y = (2x^2)/(4x^2 - 25)
40. y = √(x^3 - 9x)
42. y = x^2(x + 3)^4
44. y = (1 - x)^2/3
46. y = x ln x
48. y = x^-1e^x

Added by Theresa S.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Gregory Higby

Step 1

For each function, we need to find the intervals where it is increasing, decreasing, and the locations of any relative extrema. Show more…

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In Problems 9-52, determine where the function is (a) increasing, (b) decreasing, and (c) determine where relative extrema occur. Do not sketch the graph. 16. y = 3 + 12x - x^2 38. y = 4x^2 - 25 18. y = x^3 - 6x^2 + 12x - 6 40. y = x^3 - 9x 30. y = x^2 - 2 42. y = x^2 + 34 44. y = 1 - x^(2/3) 46. y = x ln x 52. y = x + e^(-x) 48. y = x - e^x