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Sketch the graph of f(x) = 6xe*. A. The domain of fis R. B. The X- and y-intercepts are both 0 x- C. Symmetry: None. D. Because both 6x and e* become large as x→ ∞, we have lim 6xex=∞. As x→∞, however, ex→→, so we have an indeterminate product that requires the use of l'Hospital's Rule. e* lim 6xe* = F. X→-00 đ 6x H lim DO ĐẶ lim -6e* = E. f'(x) = 6xe* + 6e* = Thus the x-axis is a horizontal asymptote. Because f'(-1) = lim x →-00 Since e* is always positive, we see that f'(x) > 0 when x + 1 > 0, and f'(x) < 0 when x + 1 < 0. So f is increasing on G. f(x) = (6 + 6x)e* + 6e* = Since f"(x) > 0 if x > 6 and f' changes from negative to positive at x = and f"(x) < 0 if x < ,00) and decreasing on (-∞, f(-1)-6e-¹ is a local (and absolute) minimum. f is concave upward on the interval and concave downward on the interval O). The inflection point is (x, y) =

Sketch the graph of f(x) = 6xe*. A. The domain of fis R. B. The X- and y-intercepts are both 0 x- C. Symmetry: None. D. Because both 6x and e* become large as x→ ∞, we have lim 6xex=∞. As x→∞, however, ex→→, so we have an indeterminate product that requires the use of l'Hospital's Rule. e* lim 6xe* = F. X→-00 đ 6x H lim DO ĐẶ lim -6e* = E. f'(x) = 6xe* + 6e* = Thus the x-axis is a horizontal asymptote. Because f'(-1) = lim x →-00 Since e* is always positive, we see that f'(x) > 0 when x + 1 > 0, and f'(x) < 0 when x + 1 < 0. So f is increasing on G. f(x) = (6 + 6x)e* + 6e* = Since f"(x) > 0 if x > 6 and f' changes from negative to positive at x = and f"(x) < 0 if x < ,00) and decreasing on (-∞, f(-1)-6e-¹ is a local (and absolute) minimum. f is concave upward on the interval and concave downward on the interval O). The inflection point is (x, y) =

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.5: Rational Functions
Problem 64E
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Sketch the graph of f(x) A. The domain of f is R. B. C. D. E. F. G. Symmetry: None. The x- and y-intercepts are both 0 lim 6xex Because both 6x and ex become large as x →∞, we have lim 6xe* = ∞o. As x → -∞, x → ∞0 X→-∞ = = lim = X 6xex. X-80 f'(x) = 6xe* + 6e* lim -6ex = Because f'(-1) = = 6x Thus the x-axis is a horizontal asymptote. Since f"(x) > 0 if x > H = f"(x) = (6 + 6x)ex + 6e* lim X→-∞ = Since ex is always positive, we see that f'(x) > 0 when x + 1 > 0, and f'(x) < 0 when x + 1 < 0. So f is increasing on and f' changes from negative to positive at x = 6 and f'(x) < 0 if x < however, ex I f(-1): = -6e so we have an indeterminate product that requires the use of l'Hospital's Rule. I f is concave upward on the interval I ∞ and decreasing on -1 is a local (and absolute) minimum. and concave downward on the interval -8, The inflection point is (x, y) =
Transcribed Image Text:Sketch the graph of f(x) A. The domain of f is R. B. C. D. E. F. G. Symmetry: None. The x- and y-intercepts are both 0 lim 6xex Because both 6x and ex become large as x →∞, we have lim 6xe* = ∞o. As x → -∞, x → ∞0 X→-∞ = = lim = X 6xex. X-80 f'(x) = 6xe* + 6e* lim -6ex = Because f'(-1) = = 6x Thus the x-axis is a horizontal asymptote. Since f"(x) > 0 if x > H = f"(x) = (6 + 6x)ex + 6e* lim X→-∞ = Since ex is always positive, we see that f'(x) > 0 when x + 1 > 0, and f'(x) < 0 when x + 1 < 0. So f is increasing on and f' changes from negative to positive at x = 6 and f'(x) < 0 if x < however, ex I f(-1): = -6e so we have an indeterminate product that requires the use of l'Hospital's Rule. I f is concave upward on the interval I ∞ and decreasing on -1 is a local (and absolute) minimum. and concave downward on the interval -8, The inflection point is (x, y) =
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