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

Step2: Find the first - derivative and critical points

  • First - derivative: Using the power rule ((x^{n})^\prime=nx^{n - 1}), (y^\prime=-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) (in ((-\infty,0))): (y^\prime=-(-1)^{2}+2(-1)=-1 - 2=-3<0). For (x = 1) (in ((0,2))): (y^\prime=-1^{2}+2\times1=1>0). For (x = 3) (in ((2,\infty))): (y^\prime=-3^{2}+2\times3=-9 + 6=-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). Using the power rule, (y^{\prime\prime}=-2x + 2).
  • Inflection points: Set (y^{\prime\prime}=0), so (-2x + 2=0). Solving for (x) gives (x = 1).

Step5: Determine intervals of concavity

  • Test intervals: Choose test points in the intervals ((-\infty,1)) and ((1,\infty)). For (x = 0) (in ((-\infty,1))): (y^{\prime\prime}=-2\times0+2=2>0). For (x = 2) (in ((1,\infty))): (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 ((x=-1,y^\prime=-3)) to positive ((x = 1,y^\prime=1)) at (x = 0), (y(0)=0) is a relative minimum. Since (y^\prime) changes from positive ((x = 1,y^\prime=1)) to negative ((x = 3,y^\prime=-3)) at (x = 2), (y(2)=-\frac{2^{3}}{3}+2^{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) - coordinates): (x = 0) and (x = 2)
  • Intervals of increase: ((0,2))
  • Intervals of decrease: ((-\infty,0)\cup(2,\infty))
  • Inflection point ((x) - coordinate): (x = 1)
  • Intervals of 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})