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

for each problem, find the: x and y intercepts, x - coordinates of the critical points, open intervals wher the function is increasing and decreasing, x - coordinates of the inflection points, open intervals where t function is concave up and concave down, and relative minima and maxima. using this information, sketch the graph of the function. 1) ( y = -\frac{x^{3}}{3}+x^{2} )

for each problem, find the: x and y intercepts, x - coordinates of the critical points, open intervals wher the function is increasing and decreasing, x - coordinates of the inflection points, open intervals where t function is concave up and concave down, and relative minima and maxima. using this information, sketch the graph of the function. 1) ( 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}\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). Set (y^\prime = 0): (-x^{2}+2x=0). Factor out (-x): (-x(x - 2)=0). The critical points are (x = 0) and (x = 2).
  • Intervals of increase and decrease: Test intervals: For (x\lt0), let (x=-1), then (y^\prime=-(-1)^{2}+2(-1)=-1 - 2=-3\lt0). For (0\lt x\lt2), let (x = 1), then (y^\prime=-1^{2}+2(1)=1\gt0). For (x\gt2), let (x = 3), then (y^\prime=-3^{2}+2(3)=-9 + 6=-3\lt0). The function is decreasing on ((-\infty,0)\cup(2,\infty)) and increasing on ((0,2)).

Step3: Find the second derivative and inflection points

  • Second derivative: Differentiate (y^\prime=-x^{2}+2x) to get (y^{\prime\prime}=-2x + 2). Set (y^{\prime\prime}=0): (-2x + 2=0), so (x = 1).
  • Intervals of concavity: For (x\lt1), let (x = 0), then (y^{\prime\prime}=-2(0)+2=2\gt0). For (x\gt1), let (x = 2), then (y^{\prime\prime}=-2(2)+2=-2\lt0). The function is concave up on ((-\infty,1)) and concave down on ((1,\infty)).

Step4: 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)-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 concave up: ((-\infty,1))
  • Intervals of concave down: ((1,\infty))
  • Relative minimum: At (x = 0), (y = 0)
  • Relative maximum: 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).
  • Mark the inflection point ((1,y(1))) where (y(1)=-\frac{1}{3}+1=\frac{2}{3}).
  • Use the intervals of increase/decrease and concavity to shape the curve. The function is decreasing on ((-\infty,0)), increasing on ((0,2)), decreasing on ((2,\infty)), concave up on ((-\infty,1)) and concave down on ((1,\infty)).