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. 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 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. 1) $y = -\frac{x^3}{3} + x^2$

Answer

Explanation:

Step1: Find x and y intercepts

To find the y - intercept, set (x = 0) in (y=-\frac{x^{3}}{3}+x^{2}). Then (y = 0), so the y - intercept is ((0,0)).

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

Step2: Find critical points (derivative = 0)

First, find the first derivative (y^\prime). Using the power rule, if (y =-\frac{x^{3}}{3}+x^{2}), then (y^\prime=-x^{2}+2x).

Set (y^\prime = 0): [ \begin{align*} -x^{2}+2x&=0\ x(-x + 2)&=0 \end{align*} ] So (x = 0) or (x = 2). These are the x - coordinates of the critical points.

Step3: Intervals of increasing/decreasing

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). (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). (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). (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 (left of (x = 0)) to increasing (right of (x = 0))? Wait, no. Wait, left of (x = 0) ((x<0)) the function is decreasing, right of (x = 0) (between (0) and (2)) the function is increasing. Wait, no, when (x) moves from left of (0) to right of (0), the derivative changes from negative to positive? Wait, no, at (x=-1) (left of (0)) (y^\prime=-3<0), at (x = 1) (right of (0) and left of (2)) (y^\prime = 1>0). Wait, actually, at (x = 0), the function has a relative minimum? Wait, no, wait: when (x) is less than (0), the function is decreasing, when (x) is between (0) and (2), the function is increasing. So at (x = 0), the function has a relative minimum.

At (x = 2), the function changes from increasing (left of (2)) to decreasing (right of (2)). So at (x = 2), the function has a relative maximum.

To find the y - values of the relative extrema:

  • For (x = 0), (y=-\frac{0^{3}}{3}+0^{2}=0), so the relative minimum is ((0,0)).
  • For (x = 2), (y=-\frac{2^{3}}{3}+2^{2}=-\frac{8}{3}+4=\frac{-8 + 12}{3}=\frac{4}{3}), so the relative maximum is ((2,\frac{4}{3})).

Step5: Find inflection points (second derivative = 0)

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

Set (y^{\prime\prime}=0): [ \begin{align*} -2x + 2&=0\ -2x&=-2\ x&=1 \end{align*} ] The x - coordinate of the inflection point is (x = 1).

Step6: Intervals of concavity

We use (x = 1) to divide the real line into intervals: ((-\infty,1)) and ((1,\infty)).

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

To find the y - value of the inflection point, substitute (x = 1) into the original function: (y=-\frac{1^{3}}{3}+1^{2}=-\frac{1}{3}+1=\frac{2}{3}). So the inflection point is ((1,\frac{2}{3})).

Answer:

  • x - intercepts: ((0,0)), ((3,0))
  • y - intercept: ((0,0))
  • x - coordinates of critical points: (x = 0), (x = 2)
  • Intervals of increasing: ((0,2))
  • Intervals of decreasing: ((-\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}))
  • Inflection point: ((1,\frac{2}{3}))

To sketch the graph:

  • Plot the intercepts ((0,0)) and ((3,0)), the relative extrema ((0,0)) and ((2,\frac{4}{3})), and 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)), and concave down on ((1,\infty)). Connect the points according to these behaviors.