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

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) in the function (y=-\frac{0^{3}}{3}+0^{2}=0). So the y - intercept is ((0,0)).

Step2: Find the first derivative (for critical points and increasing/decreasing)

The function is (y =-\frac{x^{3}}{3}+x^{2}). Using the power rule ((x^{n})^\prime=nx^{n - 1}), the first derivative (y^\prime) is: [ y^\prime=-x^{2}+2x ] To find critical points, set (y^\prime = 0): [ -x^{2}+2x=0 ] Factor out (-x): [ -x(x - 2)=0 ] This gives (x = 0) or (x=2). These are the x - coordinates of the critical points.

Step3: Determine increasing and decreasing intervals

We use test points in the intervals ((-\infty,0)), ((0,2)) and ((2,\infty)).

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

Step4: Find the second derivative (for inflection points and concavity)

The first derivative is (y^\prime=-x^{2}+2x). The second derivative (y^{\prime\prime}) is: [ y^{\prime\prime}=-2x + 2 ] To find inflection points, set (y^{\prime\prime}=0): [ -2x + 2=0 ] Solve for (x): (2x=2) so (x = 1).

Step5: Determine concavity intervals

We use test points in the intervals ((-\infty,1)) and ((1,\infty)).

  • For the interval ((-\infty,1)), let's choose (x = 0). Then (y^{\prime\prime}=-2(0)+2=2>0). So the function is concave up on ((-\infty,1)).
  • For the interval ((1,\infty)), let's choose (x = 2). Then (y^{\prime\prime}=-2(2)+2=-4 + 2=-2<0). So the function is concave down on ((1,\infty)).

Step6: Find relative minima and maxima

We use the first derivative test.

  • At (x = 0): The function changes from decreasing (on ((-\infty,0))) to increasing (on ((0,2)))? No, wait: on ((-\infty,0)) it is decreasing, on ((0,2)) it is increasing. Wait, no: when (x) approaches 0 from the left, the function is decreasing, and when (x) approaches 0 from the right, the function is increasing? Wait, no, at (x = 0), the left interval ((-\infty,0)) is decreasing, right interval ((0,2)) is increasing. Wait, no, the sign of (y^\prime) changes from negative (left of 0) to positive (right of 0)? Wait, no, at (x=-1) (left of 0), (y^\prime=-3<0), at (x = 1) (right of 0), (y^\prime = 1>0). So at (x = 0), the function has a relative minimum? Wait, no, wait: when the derivative changes from negative to positive, the function has a relative minimum. When it changes from positive to negative, it has a relative maximum.

At (x = 0): Left of (x = 0), (y^\prime<0) (decreasing), right of (x = 0), (y^\prime>0) (increasing). So (x = 0) is a relative minimum. The value of the function at (x = 0) is (y(0)=0).

At (x = 2): Left of (x = 2) (e.g., (x = 1)), (y^\prime=1>0) (increasing), right of (x = 2) (e.g., (x = 3)), (y^\prime=-3<0) (decreasing). So (x = 2) is a relative maximum. The value of the function at (x = 2) is (y(2)=-\frac{8}{3}+4=\frac{-8 + 12}{3}=\frac{4}{3}).

Answer:

  • x - intercepts: ((0,0)), ((3,0))
  • y - intercept: ((0,0))
  • x - coordinates of critical points: (x = 0), (x = 2)
  • Increasing intervals: ((0,2))
  • Decreasing intervals: ((-\infty,0)), ((2,\infty))
  • x - coordinate of inflection point: (x = 1)
  • Concave up interval: ((-\infty,1))
  • Concave down interval: ((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)).
  • Plot the critical points ((0,0)) (relative minimum) and ((2,\frac{4}{3})) (relative maximum).
  • Plot the inflection point ((1,y(1))). (y(1)=-\frac{1}{3}+1=\frac{2}{3}), so ((1,\frac{2}{3})).
  • The graph is decreasing on ((-\infty,0)), increasing on ((0,2)), decreasing on ((2,\infty)), concave up on ((-\infty,1)), concave down on ((1,\infty)). Connect the points according to the increasing/decreasing and concavity information.