Example 1 Video Example-4 Use the guidelines to sketch the curve y A. The domain is...

Transcribed Image Text:

Example 1 Video Example-4 Use the guidelines to sketch the curve y A. The domain is (x1x²-10) (*1** *1) B. The x- and y-intercepts are both 0. C. Since -x) - fx), the function is elec D. lim 112 Therefore, the line y lim *-1* 2x² lim 4-1²-1 2x² F421 tim 2x² 2 lim 7-11- lim (using interval notation). M Since the denominator is 0 when x 1, we compute the following limits. 2x² The curve is symmetric about the y-axis. is a horizontal asymptote. MY NOTES ASK YOUR therefore, the ines x 1 and 4². 2x²-2x E. (). Since F'(x) > 0 when x<0 (x-1) and F'(x) < 0 when 0(x+1), Fis increasing on the intervals (-,-1) and G. (1)- F. The only critical number is x-0. Since changes from positive to negative at -4(x²-1)² + 4x20x²-1)2x are vertical asymptotes. This information about limits and asymptotes enables us to draw the asymptotes in the following figure Since the simplified numerator, w >0 for all x, we have the following. and x) <0. F(x)0²-10 1x1 Thus, the curve is concave upward on the intervals (-, -1) and domain off 1x1 < and concave downward on and decreasing on the interes ( 13 and is a local maximum by the First Derivative Test It has no point of inflection since 1 and 2 are not in the H. Using the information in parts (E)-(G), we finish the sketch in the following figure. y = 2 1 Need Help? Read It Use the guidelines to sketch the curve y A. The domain is (x1x²-10) (x1x* *1) B. The x- and y-intercepts are both 0. D. C Since f(-x) x), the function fis-Select- The curve is symmetric about the y-axis. lim - lim 2x² lim x-12 lim 11' lim Therefore, the line y is a horizontal asymptote Since the denominator is 0 when xa1, we compute the following limits. 24² lim *1 2x² (using interval notation). " Therefore, the lines x = 1 and x-[ 8.) 4(x²-1)-2²-2 (x²-1)² Since /(x) > 0 when x < 0(x-1) and (x) <0 when x>0 (x1), fis increasing on the intervals (-,-1) and are vertical asymptotes. This information about limits and asymptotes enables us to draw the asymptotes in the following figure. F. The only critical number is x = 0. Since / changes from positive to negative at G. x)=-4(x²-1)² + 4x2(x²-132x (²-1) Since the simplified numerator 10)>0x²10 > > 0 for all x, we have the following. and f(x) < < Thus, the curve is concave upward on the intervals (-, -1) and domain off and concave downward on and decreasing on the intervals (0, 1) and is a local maximum by the First Derivative Test. It has no point of inflection since 1 and are not in the H. Using the information in parts (E)-(G), we finish the sketch in the following figure. y = 2 x=- Example 1 Video Example-4 Use the guidelines to sketch the curve y A. The domain is (x1x²-10) (*1** *1) B. The x- and y-intercepts are both 0. C. Since -x) - fx), the function is elec D. lim 112 Therefore, the line y lim *-1* 2x² lim 4-1²-1 2x² F421 tim 2x² 2 lim 7-11- lim (using interval notation). M Since the denominator is 0 when x 1, we compute the following limits. 2x² The curve is symmetric about the y-axis. is a horizontal asymptote. MY NOTES ASK YOUR therefore, the ines x 1 and 4². 2x²-2x E. (). Since F'(x) > 0 when x<0 (x-1) and F'(x) < 0 when 0(x+1), Fis increasing on the intervals (-,-1) and G. (1)- F. The only critical number is x-0. Since changes from positive to negative at -4(x²-1)² + 4x20x²-1)2x are vertical asymptotes. This information about limits and asymptotes enables us to draw the asymptotes in the following figure Since the simplified numerator, w >0 for all x, we have the following. and x) <0. F(x)0²-10 1x1 Thus, the curve is concave upward on the intervals (-, -1) and domain off 1x1 < and concave downward on and decreasing on the interes ( 13 and is a local maximum by the First Derivative Test It has no point of inflection since 1 and 2 are not in the H. Using the information in parts (E)-(G), we finish the sketch in the following figure. y = 2 1 Need Help? Read It Use the guidelines to sketch the curve y A. The domain is (x1x²-10) (x1x* *1) B. The x- and y-intercepts are both 0. D. C Since f(-x) x), the function fis-Select- The curve is symmetric about the y-axis. lim - lim 2x² lim x-12 lim 11' lim Therefore, the line y is a horizontal asymptote Since the denominator is 0 when xa1, we compute the following limits. 24² lim *1 2x² (using interval notation). " Therefore, the lines x = 1 and x-[ 8.) 4(x²-1)-2²-2 (x²-1)² Since /(x) > 0 when x < 0(x-1) and (x) <0 when x>0 (x1), fis increasing on the intervals (-,-1) and are vertical asymptotes. This information about limits and asymptotes enables us to draw the asymptotes in the following figure. F. The only critical number is x = 0. Since / changes from positive to negative at G. x)=-4(x²-1)² + 4x2(x²-132x (²-1) Since the simplified numerator 10)>0x²10 > > 0 for all x, we have the following. and f(x) < < Thus, the curve is concave upward on the intervals (-, -1) and domain off and concave downward on and decreasing on the intervals (0, 1) and is a local maximum by the First Derivative Test. It has no point of inflection since 1 and are not in the H. Using the information in parts (E)-(G), we finish the sketch in the following figure. y = 2 x=-

Answer rating: 100% (QA)

Here are the stepbystep workings A The domain is all real numbers except x 1 Using interval notation ... View the full answer