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 - intercepts

Set (y = 0), so (0=-\frac{x^{3}}{3}+x^{2}). Factor out (x^{2}): (x^{2}( -\frac{x}{3}+ 1)=0). Then (x = 0) or (x = 3).

Step2: Find y - intercepts

Set (x = 0), then (y=-\frac{0^{3}}{3}+0^{2}=0).

Step3: Find the first - derivative

Differentiate (y =-\frac{x^{3}}{3}+x^{2}) with respect to (x). Using the power rule ((x^{n})^\prime=nx^{n - 1}), (y^\prime=-x^{2}+2x).

Step4: Find 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).

Step5: Determine intervals of increase and decrease

Test intervals: For (x\lt0), let (x=-1), then (y^\prime=-(-1)^{2}+2(-1)=-1 - 2=-3\lt0), so the function is decreasing on ((-\infty,0)). For (0\lt x\lt2), let (x = 1), then (y^\prime=-1^{2}+2\times1=1\gt0), so the function is increasing on ((0,2)). For (x\gt2), let (x = 3), then (y^\prime=-3^{2}+2\times3=-9 + 6=-3\lt0), so the function is decreasing on ((2,\infty)).

Step6: Find the second - derivative

Differentiate (y^\prime=-x^{2}+2x) with respect to (x). (y^{\prime\prime}=-2x + 2).

Step7: Find inflection points

Set (y^{\prime\prime}=0), so (-2x + 2=0), which gives (x = 1).

Step8: Determine concavity

For (x\lt1), let (x = 0), then (y^{\prime\prime}=-2\times0+2 = 2\gt0), so the function is concave up on ((-\infty,1)). For (x\gt1), let (x = 2), then (y^{\prime\prime}=-2\times2+2=-2\lt0), so the function is concave down on ((1,\infty)).

Step9: Find relative extrema

Since the function changes from decreasing ((x\lt0)) to increasing ((0\lt x\lt2)) at (x = 0), (y(0)=0) is a relative minimum. Since the function changes from increasing ((0\lt x\lt2)) to decreasing ((x\gt2)) 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,x = 3)
  • y - intercept: (y = 0)
  • Critical points: (x = 0,x = 2)
  • Increasing intervals: ((0,2))
  • Decreasing intervals: ((-\infty,0)\cup(2,\infty))
  • Inflection point: (x = 1)
  • Concave - up interval: ((-\infty,1))
  • Concave - down interval: ((1,\infty))
  • Relative minimum: ((0,0))
  • Relative maximum: ((2,\frac{4}{3}))