for each problem, find the: x and y intercepts, x - coordinates of the critical points, open intervals where…

for each problem, find the: x and y intercepts, x - coordinates of the critical points, open intervals where the function is increasing and decreasing, x - coordinates of the inflection points, open intervals where the function is concave up and concave down, and relative minima and maxima. using this information, sketch the graph of the function.\n1) ( y =-\frac{x^{3}}{3}+x^{2} )

for each problem, find the: x and y intercepts, x - coordinates of the critical points, open intervals where the function is increasing and decreasing, x - coordinates of the inflection points, open intervals where the function is concave up and concave down, and relative minima and maxima. using this information, sketch the graph of the function.\n1) ( y =-\frac{x^{3}}{3}+x^{2} )

Answer

Explanation:

Step1: Find (x) and (y) - intercepts

  • (y)-intercept: Set (x = 0) in (y=-\frac{x^{3}}{3}+x^{2}). Then (y = 0).
  • (x)-intercept: Set (y = 0), so (0=-\frac{x^{3}}{3}+x^{2}). Factor out (x^{2}): (x^{2}(1-\frac{x}{3}) = 0). Solving (x^{2}=0) gives (x = 0), and solving (1-\frac{x}{3}=0) gives (x = 3).

Step2: Find the first - derivative and critical points

  • First - derivative: Using the power rule (y^\prime=\frac{d}{dx}(-\frac{x^{3}}{3}+x^{2})=-x^{2}+2x).
  • Critical points: Set (y^\prime = 0), so (-x^{2}+2x=0). Factor out (-x): (-x(x - 2)=0). The critical points are (x = 0) and (x = 2).

Step3: Determine intervals of increase and decrease

  • Test intervals: Choose test points in the intervals ((-\infty,0)), ((0,2)), and ((2,\infty)). For (x=-1), (y^\prime=-(-1)^{2}+2(-1)=-3<0). For (x = 1), (y^\prime=-1^{2}+2\times1 = 1>0). For (x = 3), (y^\prime=-3^{2}+2\times3=-3<0). The function is decreasing on ((-\infty,0)\cup(2,\infty)) and increasing on ((0,2)).

Step4: Find the second - derivative and inflection points

  • Second - derivative: Differentiate (y^\prime=-x^{2}+2x) with respect to (x). (y^{\prime\prime}=-2x + 2).
  • Inflection point: Set (y^{\prime\prime}=0), so (-2x + 2=0). Solving for (x) gives (x = 1).

Step5: Determine concavity

  • Test intervals: For (x<1) (e.g., (x = 0)), (y^{\prime\prime}=-2\times0+2 = 2>0). For (x>1) (e.g., (x = 2)), (y^{\prime\prime}=-2\times2+2=-2<0). The function is concave up on ((-\infty,1)) and concave down on ((1,\infty)).

Step6: Find relative minima and maxima

  • Use the first - derivative test: Since (y^\prime) changes from negative to positive at (x = 0), (y(0)=0) is a relative minimum. Since (y^\prime) changes from positive to negative at (x = 2), (y(2)=-\frac{8}{3}+4=\frac{4}{3}) is a relative maximum.

Answer:

  • (x)-intercepts: (x = 0) and (x = 3); (y)-intercept: (y = 0).
  • Critical points: (x = 0) and (x = 2).
  • Intervals of increase: ((0,2)); Intervals of decrease: ((-\infty,0)\cup(2,\infty)).
  • Inflection point: (x = 1).
  • Concavity: Concave up on ((-\infty,1)), concave down on ((1,\infty)).
  • Relative minima: At (x = 0), (y = 0); Relative maxima: At (x = 2), (y=\frac{4}{3}).

To sketch the graph:

  • Plot the intercepts ((0,0)) and ((3,0)).
  • Mark the critical points ((0,0)) (relative minimum) and ((2,\frac{4}{3})) (relative maximum) and the inflection point ((1,\frac{2}{3})).
  • Use the information about intervals of increase/decrease and concavity to draw a smooth curve.