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). We get (x = 0) or (x = 3).

Step2: Find the first - derivative and critical points

The function (y=-\frac{x^{3}}{3}+x^{2}). The first - derivative (y^{\prime}=-x^{2}+2x). Set (y^{\prime}=0), then (-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

  • Consider the intervals ((-\infty,0)), ((0,2)) and ((2,\infty)) Take a test point in ((-\infty,0)), say (x=-1). Then (y^{\prime}=-(-1)^{2}+2(-1)=-1 - 2=-3<0). Take a test point in ((0,2)), say (x = 1). Then (y^{\prime}=-1^{2}+2\times1=1>0). Take a test point in ((2,\infty)), say (x = 3). Then (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

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

Step5: Determine intervals of concavity

  • Consider the intervals ((-\infty,1)) and ((1,\infty)) Take a test point in ((-\infty,1)), say (x = 0). Then (y^{\prime\prime}=-2\times0+2=2>0). Take a test point in ((1,\infty)), say (x = 2). Then (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 sign from negative to positive at (x = 0), (y(0)=0) is a relative minimum. Since (y^{\prime}) changes sign from positive to negative at (x = 2), (y(2)=-\frac{2^{3}}{3}+2^{2}=-\frac{8}{3}+4=\frac{-8 + 12}{3}=\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)
  • 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})