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) $y = -\\dfrac{x^3}{3} + x^2$

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) $y = -\\dfrac{x^3}{3} + x^2$

Answer

Explanation:

Step1: Find x and y intercepts

To find the x - intercepts, set (y = 0): [ -\frac{x^{3}}{3}+x^{2}=0 ] Factor out (x^{2}): [ x^{2}\left(-\frac{x}{3} + 1\right)=0 ] This gives (x^{2}=0) or (-\frac{x}{3}+1 = 0). Solving (x^{2}=0) gives (x = 0). Solving (-\frac{x}{3}+1=0) gives (x=3). So the x - intercepts are ((0,0)) and ((3,0)).

To find the y - intercept, set (x = 0): [ y=-\frac{0^{3}}{3}+0^{2}=0 ] So the y - intercept is ((0,0)).

Step2: Find critical points (derivative of the function)

First, find the first derivative (y^\prime) of (y =-\frac{x^{3}}{3}+x^{2}). Using the power rule ((x^{n})^\prime=nx^{n - 1}), we have: [ y^\prime=-x^{2}+2x ] Set (y^\prime = 0) to find critical points: [ -x^{2}+2x=0 ] Factor out (-x): [ -x(x - 2)=0 ] This gives (-x=0) (i.e., (x = 0)) or (x - 2=0) (i.e., (x = 2)). So the x - coordinates of critical points are (x = 0) and (x = 2).

Step3: Intervals of increase and decrease

We use the critical points (x = 0) and (x = 2) to divide the real line into intervals ((-\infty,0)), ((0,2)), and ((2,\infty)).

  • For (x\in(-\infty,0)), let's test (x=-1) in (y^\prime=-x^{2}+2x). Then (y^\prime=-(-1)^{2}+2(-1)=-1 - 2=-3<0). So the function is decreasing on ((-\infty,0)).
  • For (x\in(0,2)), let's test (x = 1) in (y^\prime=-x^{2}+2x). Then (y^\prime=-1^{2}+2(1)=-1 + 2 = 1>0). So the function is increasing on ((0,2)).
  • For (x\in(2,\infty)), let's test (x = 3) in (y^\prime=-x^{2}+2x). Then (y^\prime=-3^{2}+2(3)=-9 + 6=-3<0). So the function is decreasing on ((2,\infty)).

Step4: Find relative minima and maxima

Using the first - derivative test:

  • At (x = 0), the function changes from decreasing to increasing? No, wait, from ((-\infty,0)) (decreasing) to ((0,2)) (increasing). Wait, when (x) moves from left of (0) to right of (0), the derivative changes from negative to positive? Wait, no, at (x = 0), left of (0) (e.g., (x=-1)) (y^\prime<0), right of (0) (e.g., (x = 1)) (y^\prime>0). Wait, no, my mistake earlier: when (x\in(-\infty,0)), (y^\prime<0) (decreasing), when (x\in(0,2)), (y^\prime>0) (increasing). So at (x = 0), the function has a relative minimum. The value of the function at (x = 0) is (y(0)=0), so relative minimum at ((0,0)).

  • At (x = 2), the function changes from increasing (on ((0,2))) to decreasing (on ((2,\infty))). So the function has a relative maximum at (x = 2). The value of the function at (x = 2) is (y(2)=-\frac{2^{3}}{3}+2^{2}=-\frac{8}{3}+4=\frac{4}{3}), so relative maximum at ((2,\frac{4}{3})).

Step5: Find the second derivative (for inflection points and concavity)

First, find the second derivative (y^{\prime\prime}) of (y). Since (y^\prime=-x^{2}+2x), then (y^{\prime\prime}=-2x + 2).

Step6: Find inflection points

Set (y^{\prime\prime}=0): [ -2x + 2=0 ] Solve for (x): [ -2x=-2\implies x = 1 ] To confirm it's an inflection point, we check the sign of (y^{\prime\prime}) on either side of (x = 1).

  • For (x<1) (e.g., (x = 0)), (y^{\prime\prime}=-2(0)+2=2>0).
  • For (x>1) (e.g., (x = 2)), (y^{\prime\prime}=-2(2)+2=-2<0). So (x = 1) is the x - coordinate of the inflection point. The value of the function at (x = 1) is (y(1)=-\frac{1}{3}+1=\frac{2}{3}), so the inflection point is ((1,\frac{2}{3})).

Step7: Intervals of concavity

  • When (y^{\prime\prime}>0), the function is concave up. (y^{\prime\prime}=-2x + 2>0\implies -2x>-2\implies x<1). So the function is concave up on ((-\infty,1)).
  • When (y^{\prime\prime}<0), the function is concave down. (y^{\prime\prime}=-2x + 2<0\implies -2x<-2\implies x>1). So the function is concave down on ((1,\infty)).

Answer:

  • x - intercepts: ((0,0)), ((3,0))
  • y - intercept: ((0,0))
  • x - coordinates of critical points: (x = 0), (x = 2)
  • Intervals of increase: ((0,2))
  • Intervals of decrease: ((-\infty,0)), ((2,\infty))
  • x - coordinate of inflection point: (x = 1)
  • Intervals of concave up: ((-\infty,1))
  • Intervals of concave down: ((1,\infty))
  • Relative minimum: ((0,0))
  • Relative maximum: ((2,\frac{4}{3}))

(For sketching the graph: Plot the intercepts ((0,0)) and ((3,0)), the relative minimum ((0,0)), the relative maximum ((2,\frac{4}{3})), the inflection point ((1,\frac{2}{3})). The function is decreasing on ((-\infty,0)), increasing on ((0,2)), decreasing on ((2,\infty)), concave up on ((-\infty,1)), concave down on ((1,\infty)).)