for each problem, find the: x and y intercepts, x - coordinates of the critical points, open intervals wher…

for each problem, find the: x and y intercepts, x - coordinates of the critical points, open intervals wher the function is increasing and decreasing, x - coordinates of the inflection points, open intervals where t 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 wher the function is increasing and decreasing, x - coordinates of the inflection points, open intervals where t 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 (set ( y = 0 ))

( 0 = -\frac{x^3}{3}+x^2 )
Factor: ( x^2\left(1 - \frac{x}{3}\right)=0 )
Solutions: ( x = 0 ) or ( x = 3 ). So x-intercepts at ( (0, 0) ), ( (3, 0) ).

Step2: Find y-intercept (set ( x = 0 ))

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

Step3: Find critical points (derivative ( y' = 0 ))

( y' = -x^2 + 2x )
Set ( -x^2 + 2x = 0 )
Factor: ( x(-x + 2) = 0 )
Solutions: ( x = 0 ), ( x = 2 ).

Step4: Intervals of increase/decrease (test ( y' ))

  • For ( x < 0 ) (e.g., ( x = -1 )): ( y' = -1 - 2 = -3 < 0 ) → decreasing.
  • For ( 0 < x < 2 ) (e.g., ( x = 1 )): ( y' = -1 + 2 = 1 > 0 ) → increasing.
  • For ( x > 2 ) (e.g., ( x = 3 )): ( y' = -9 + 6 = -3 < 0 ) → decreasing.

Step5: Relative extrema (using sign changes of ( y' ))

  • At ( x = 0 ): ( y' ) doesn’t change sign (both sides negative/positive? Wait, left: decreasing, right: increasing? Wait, no: left of 0: decreasing, right (0 to 2): increasing. Wait, sign changes from - to +? Wait, no: ( x < 0 ): ( y' < 0 ); ( 0 < x < 2 ): ( y' > 0 ). So at ( x = 0 ), ( y' ) changes from - to +? Wait, no: ( x = -1 ): ( y' = -(-1)^2 + 2(-1) = -1 - 2 = -3 ); ( x = 1 ): ( y' = -1 + 2 = 1 ). So at ( x = 0 ), left: negative, right: positive → local minimum? Wait, no: wait, ( x = 0 ): when moving from left to right, ( y' ) goes from - to + → local minimum? Wait, but ( x = 2 ): left (0 to 2): ( y' > 0 ), right (x > 2): ( y' < 0 ) → local maximum.

Wait, recalculate ( y' ) at ( x = 0 ): ( y' = -0 + 0 = 0 ). At ( x = 2 ): ( y' = -4 + 4 = 0 ).

So:

  • ( x = 0 ): left (decreasing), right (increasing) → local minimum.
  • ( x = 2 ): left (increasing), right (decreasing) → local maximum.

Calculate ( y ) at extrema:

  • At ( x = 0 ): ( y = 0 ) → ( (0, 0) ) (local min? Wait, but ( x = 2 )): ( y = -\frac{8}{3}+4 = \frac{4}{3} ) → ( (2, \frac{4}{3}) ) (local max).

Step6: Find inflection points (second derivative ( y'' = 0 ))

( y'' = -2x + 2 )
Set ( -2x + 2 = 0 ) → ( x = 1 ).

Step7: Intervals of concavity (test ( y'' ))

  • For ( x < 1 ) (e.g., ( x = 0 )): ( y'' = 2 > 0 ) → concave up.
  • For ( x > 1 ) (e.g., ( x = 2 )): ( y'' = -4 + 2 = -2 < 0 ) → concave down.

Summary:

  • x-intercepts: ( (0, 0) ), ( (3, 0) )
  • y-intercept: ( (0, 0) )
  • Critical points (x-coordinates): ( x = 0 ), ( x = 2 )
  • Increasing intervals: ( (0, 2) )
  • Decreasing intervals: ( (-\infty, 0) ), ( (2, \infty) )
  • Inflection point (x-coordinate): ( x = 1 )
  • Concave up interval: ( (-\infty, 1) )
  • Concave down interval: ( (1, \infty) )
  • Relative minima: ( (0, 0) )
  • Relative maxima: ( (2, \frac{4}{3}) )

To sketch the graph:

  • Plot intercepts ( (0,0) ), ( (3,0) ).
  • Plot extrema ( (0,0) ) (local min), ( (2, \frac{4}{3}) ) (local max).
  • Plot inflection point at ( x = 1 ): ( y = -\frac{1}{3}+1 = \frac{2}{3} ) → ( (1, \frac{2}{3}) ).
  • Connect with curves: decreasing from ( -\infty ) to 0, increasing from 0 to 2, decreasing from 2 to ( \infty ); concave up before ( x = 1 ), concave down after ( x = 1 ).