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-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 )

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

( y = -\frac{0^3}{3}+0^2 = 0 )

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

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

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

  • For ( x < 0 ): Let ( x = -1 ), ( y' = -(-1)^2 + 2(-1)= -3 < 0 ) (decreasing)
  • For ( 0 < x < 2 ): Let ( x = 1 ), ( y' = -(1)^2 + 2(1)= 1 > 0 ) (increasing)
  • For ( x > 2 ): Let ( x = 3 ), ( y' = -(3)^2 + 2(3)= -3 < 0 ) (decreasing)

Step5: Inflection points (second derivative ( y'' = 0 ))

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

Step6: Concavity (test ( y'' ))

  • For ( x < 1 ): Let ( x = 0 ), ( y'' = 2 > 0 ) (concave up)
  • For ( x > 1 ): Let ( x = 2 ), ( y'' = -2 < 0 ) (concave down)

Step7: Relative extrema (use first derivative test)

  • At ( x = 0 ): Left decreasing, right increasing? No, left decreasing, right increasing? Wait, left of ( x = 0 ): decreasing, right (0 to 2): increasing → ( x = 0 ) is a relative minimum (value: ( y(0) = 0 ))
  • At ( x = 2 ): Left increasing, right decreasing → relative maximum (value: ( y(2)= -\frac{8}{3}+4=\frac{4}{3} ))

Answer:

  • x-intercepts: ( x = 0 ), ( x = 3 )
  • y-intercept: ( y = 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 minimum: At ( x = 0 ), ( y = 0 )
  • Relative maximum: At ( x = 2 ), ( y = \frac{4}{3} )

(For graph sketching: Plot intercepts (0,0), (3,0); critical points (0,0) [min], (2, 4/3) [max]; inflection point (1, y(1)= -1/3 + 1 = 2/3). Use concavity and increase/decrease to draw the curve.)