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 (y = 0)

Set ( y = 0 ), so ( -\frac{x^3}{3}+x^2 = 0 ). Factor out ( x^2 ): ( x^2\left(-\frac{x}{3}+1\right)=0 ). Solve ( x^2 = 0 ) (gives ( x = 0 )) and ( -\frac{x}{3}+1 = 0 ) (gives ( x = 3 )). So x-intercepts at ( x = 0 ) and ( x = 3 ), points ( (0,0) ) and ( (3,0) ).

Step2: Find y-intercept (x = 0)

Substitute ( x = 0 ) into ( y = -\frac{0^3}{3}+0^2 = 0 ). So y-intercept at ( (0,0) ).

Step3: Find critical points (first derivative)

Differentiate ( y = -\frac{x^3}{3}+x^2 ). ( y' = -x^2 + 2x ). Set ( y' = 0 ): ( -x^2 + 2x = 0 ), factor: ( x(-x + 2)=0 ). Solutions ( x = 0 ) and ( x = 2 ).

Step4: Intervals of increase/decrease

Test intervals:

  • ( (-\infty, 0) ): Pick ( x = -1 ), ( y' = -(-1)^2 + 2(-1)= -3 < 0 ) (decreasing).
  • ( (0, 2) ): Pick ( x = 1 ), ( y' = -1 + 2 = 1 > 0 ) (increasing).
  • ( (2, \infty) ): Pick ( x = 3 ), ( y' = -9 + 6 = -3 < 0 ) (decreasing).

Step5: Relative extrema

At ( x = 0 ): left decreasing, right increasing? Wait, no: left of 0 (decreasing), right (0,2) increasing. So ( x = 0 ) is a relative minimum? Wait, no: when ( x < 0 ), decreasing; ( 0 < x < 2 ), increasing. So ( x = 0 ) is a relative minimum (value ( y(0)=0 )). At ( x = 2 ): left increasing, right decreasing. So relative maximum. ( y(2)= -\frac{8}{3}+4 = \frac{4}{3} ), point ( (2, \frac{4}{3}) ).

Step6: Inflection points (second derivative)

Differentiate ( y' = -x^2 + 2x ) to get ( y'' = -2x + 2 ). Set ( y'' = 0 ): ( -2x + 2 = 0 ) → ( x = 1 ).

Step7: Concavity intervals

Test intervals:

  • ( (-\infty, 1) ): Pick ( x = 0 ), ( y'' = 2 > 0 ) (concave up).
  • ( (1, \infty) ): Pick ( x = 2 ), ( y'' = -2 < 0 ) (concave down). So inflection point at ( x = 1 ), ( y(1)= -\frac{1}{3}+1 = \frac{2}{3} ), point ( (1, \frac{2}{3}) ).

Step8: Sketching

  • Plot intercepts ( (0,0) ), ( (3,0) ).
  • Plot extrema: ( (0,0) ) (min), ( (2, \frac{4}{3}) ) (max).
  • Plot inflection point ( (1, \frac{2}{3}) ).
  • Draw curve: decreasing on ( (-\infty,0) ), increasing on ( (0,2) ), decreasing on ( (2,\infty) ); concave up on ( (-\infty,1) ), concave down on ( (1,\infty) ).

Answer:

  • x-intercepts: ( (0, 0) ), ( (3, 0) )
  • y-intercept: ( (0, 0) )
  • Critical points: ( x = 0 ), ( x = 2 )
  • Intervals of increase: ( (0, 2) ); decrease: ( (-\infty, 0) ), ( (2, \infty) )
  • Relative minimum: ( (0, 0) ); relative maximum: ( \left(2, \frac{4}{3}\right) )
  • Inflection point: ( \left(1, \frac{2}{3}\right) )
  • Concavity up: ( (-\infty, 1) ); concavity down: ( (1, \infty) )

(Graph sketch: Plot the points, follow increase/decrease and concavity.)