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 th 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 th 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). Using the zero - product property (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: Differentiate (y=-\frac{x^{3}}{3}+x^{2}) using the power rule (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: The critical points divide the real line into intervals ((-\infty,0)), ((0,2)), and ((2,\infty)).
  • For (x=-1) (in ((-\infty,0))): (y^\prime=-(-1)^{2}+2(-1)=-1 - 2=-3<0), so the function is decreasing on ((-\infty,0)).
  • For (x = 1) (in ((0,2))): (y^\prime=-1^{2}+2\times1=1>0), so the function is increasing on ((0,2)).
  • For (x = 3) (in ((2,\infty))): (y^\prime=-3^{2}+2\times3=-9 + 6=-3<0), so the function is decreasing on ((2,\infty)).

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). Solve for (x): (2x=2), (x = 1).

Step5: Determine intervals of concavity

  • Test intervals: The inflection point (x = 1) divides the real line into intervals ((-\infty,1)) and ((1,\infty)).
  • For (x = 0) (in ((-\infty,1))): (y^{\prime\prime}=-2\times0+2=2>0), so the function is concave up on ((-\infty,1)).
  • For (x = 2) (in ((1,\infty))): (y^{\prime\prime}=-2\times2+2=-2<0), so the function is concave down on ((1,\infty)).

Step6: Find relative minima and maxima

  • Use the first - derivative test: Since the function changes from decreasing ((-\infty,0)) to increasing ((0,2)), at (x = 0): (y=-\frac{0^{3}}{3}+0^{2}=0) (a relative minimum). Since the function changes from increasing ((0,2)) to decreasing ((2,\infty)), at (x = 2): (y=-\frac{2^{3}}{3}+2^{2}=-\frac{8}{3}+4=\frac{-8 + 12}{3}=\frac{4}{3}) (a relative maximum).

Answer:

  • (x)-intercepts: (x = 0) and (x = 3); (y)-intercept: (y = 0).
  • Critical points: (x = 0) and (x = 2).
  • Increasing interval: ((0,2)); Decreasing intervals: ((-\infty,0)) and ((2,\infty)).
  • Inflection point: (x = 1).
  • Concave up interval: ((-\infty,1)); Concave down interval: ((1,\infty)).
  • Relative minimum: ((0,0)); Relative maximum: (\left(2,\frac{4}{3}\right)).