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

  • x-intercepts: Set ( y = 0 ), so ( -\frac{x^3}{3}+x^2 = 0 ). Factor out ( x^2 ): ( x^2\left(-\frac{x}{3}+1\right)=0 ). Solutions are ( x = 0 ) and ( -\frac{x}{3}+1 = 0\Rightarrow x = 3 ). So x-intercepts at ( (0,0) ) and ( (3,0) ).
  • y-intercept: Set ( x = 0 ), then ( y = -\frac{0^3}{3}+0^2 = 0 ). So y-intercept at ( (0,0) ).

Step2: Find critical points (derivative = 0)

  • First derivative: ( y'=-\frac{3x^2}{3}+2x=-x^2 + 2x ).
  • Set ( y' = 0 ): ( -x^2 + 2x = 0\Rightarrow x(-x + 2)=0 ). Solutions: ( x = 0 ) and ( x = 2 ). These are x-coordinates of critical points.

Step3: Intervals of increasing/decreasing

  • Test intervals: ( (-\infty,0) ), ( (0,2) ), ( (2,\infty) ).
  • For ( (-\infty,0) ), pick ( x=-1 ): ( y'=-(-1)^2+2(-1)=-1 - 2=-3<0 ), so function is decreasing.
  • For ( (0,2) ), pick ( x = 1 ): ( y'=-1^2+2(1)=-1 + 2 = 1>0 ), so function is increasing.
  • For ( (2,\infty) ), pick ( x = 3 ): ( y'=-3^2+2(3)=-9 + 6=-3<0 ), so function is decreasing.

Step4: Find inflection points (second derivative = 0)

  • Second derivative: ( y''=-2x + 2 ).
  • Set ( y'' = 0 ): ( -2x + 2 = 0\Rightarrow x = 1 ).

Step5: Intervals of concavity

  • Test intervals: ( (-\infty,1) ), ( (1,\infty) ).
  • For ( (-\infty,1) ), pick ( x = 0 ): ( y''=-2(0)+2 = 2>0 ), so concave up.
  • For ( (1,\infty) ), pick ( x = 2 ): ( y''=-2(2)+2=-2<0 ), so concave down.

Step6: Relative minima and maxima

  • At ( x = 0 ): left of ( x = 0 ) function is decreasing, right is increasing? Wait, no: left of ( 0 ) (e.g., ( x=-1 )) ( y' < 0 ) (decreasing), right of ( 0 ) (e.g., ( x = 1 )) ( y' > 0 ) (increasing). Wait, no: at ( x = 0 ), left is decreasing, right is increasing? Wait, no, when ( x ) goes from ( -\infty ) to ( 0 ), function is decreasing; from ( 0 ) to ( 2 ), increasing. So at ( x = 0 ), function changes from decreasing to increasing? Wait, no, ( x = 0 ): left (decreasing), right (increasing) → relative minimum? Wait, no, wait ( x = 0 ): ( y' ) at left is negative, right is positive → so ( x = 0 ) is a relative minimum? Wait, no, wait ( x = 0 ): let's compute ( y(0)=0 ), ( y(2)=-\frac{8}{3}+4=\frac{4}{3}\approx1.333 ). Wait, at ( x = 0 ), function was decreasing before, increasing after → relative minimum? But at ( x = 2 ), function was increasing before, decreasing after → relative maximum. Wait, my mistake earlier: at ( x = 0 ), left (decreasing), right (increasing) → relative minimum. At ( x = 2 ), left (increasing), right (decreasing) → relative maximum.

Step7: Sketch the graph

  • Plot intercepts ( (0,0) ), ( (3,0) ).
  • Critical points: ( (0,0) ) (relative min), ( (2,\frac{4}{3}) ) (relative max).
  • Inflection point: ( (1,-\frac{1}{3}+1)=\left(1,\frac{2}{3}\right) ).
  • Concave up on ( (-\infty,1) ), concave down on ( (1,\infty) ).
  • Decreasing on ( (-\infty,0) ), increasing on ( (0,2) ), decreasing on ( (2,\infty) ).

Answer:

x-intercepts: ( x = 0, 3 ) (points ( (0,0) ), ( (3,0) ))

y-intercept: ( y = 0 ) (point ( (0,0) ))

Critical points (x-coordinates): ( x = 0, 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: ( (0, 0) ) (at ( x = 0 ))

Relative maximum: ( \left(2, \frac{4}{3}\right) ) (at ( x = 2 ))

(For the graph sketch, plot these key points and use the concavity/increasing-decreasing behavior to draw the curve: starts from bottom left, decreases to ( (0,0) ), increases to ( (2,\frac{4}{3}) ), then decreases to ( (3,0) ) and beyond, with inflection at ( (1,\frac{2}{3}) ) where concavity changes.)