Question

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

Explanation:

Step1: Find x-intercepts

Set \( y = 0 \): \( -\frac{x^3}{3}+x^2 = 0 \) → \( x^2(-\frac{x}{3}+1)=0 \) → \( x = 0 \) or \( x = 3 \). So x-intercepts: \( (0,0) \), \( (3,0) \).

Step2: Find y-intercept

Set \( x = 0 \): \( y = 0 \). So y-intercept: \( (0,0) \).

Step3: Find critical points (1st derivative)

\( y' = -x^2 + 2x \). Set \( y' = 0 \): \( -x^2 + 2x = 0 \) → \( x(-x + 2)=0 \) → \( x = 0 \) or \( x = 2 \). Critical x-coordinates: \( 0, 2 \).

Step4: Intervals of increase/decrease

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

Step5: Find inflection points (2nd derivative)

\( y'' = -2x + 2 \). Set \( y'' = 0 \): \( -2x + 2 = 0 \) → \( x = 1 \). Inflection x-coordinate: \( 1 \).

Step6: Concavity intervals

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

Step7: Relative minima/maxima

At \( x = 0 \): left decreasing, right increasing? No, left decreasing, right increasing? Wait, \( x = 0 \): left \( (-\infty,0) \) decreasing, right \( (0,2) \) increasing? Wait, no: \( x = 0 \), left \( y' < 0 \), right \( y' > 0 \)? Wait, no: at \( x = 0 \), left \( x=-1 \), \( y'=-3 < 0 \); right \( x = 1 \), \( y' = 1 > 0 \). Wait, so \( x = 0 \): local min? Wait, no: \( x = 0 \), function at \( x = 0 \) is \( 0 \), at \( x = 2 \): \( y = -\frac{8}{3}+4=\frac{4}{3} \). Wait, at \( x = 0 \): left decreasing, right increasing → local min? Wait, no: when \( x \) approaches \( -\infty \), \( y = -\frac{x^3}{3}+x^2 \) → \( -\infty \) (since \( -x^3/3 \) dominates). At \( x = 0 \), \( y = 0 \); at \( x = 2 \), \( y = 4/3 \); at \( x = 3 \), \( y = 0 \). Wait, at \( x = 0 \): function goes from \( -\infty \) to \( 0 \) (decreasing), then increases to \( x = 2 \), then decreases. So at \( x = 0 \): local min? Wait, no, when \( x < 0 \), function is decreasing (going to \( -\infty \)), at \( x = 0 \), then increases. Wait, actually, at \( x = 0 \), the function has a local minimum? Wait, no, let's check values: \( x = -1 \): \( y = -(-1)/3 + 1 = 1/3 + 1 = 4/3 \). Wait, wait, I miscalculated \( y \) at \( x = -1 \): \( y = -(-1)^3/3 + (-1)^2 = 1/3 + 1 = 4/3 \). Oh! So at \( x = -1 \), \( y = 4/3 \), at \( x = 0 \), \( y = 0 \), so from \( x = -\infty \) to \( x = 0 \), function decreases from \( \infty \) (wait, no: as \( x \to -\infty \), \( -x^3/3 \to \infty \) (since \( x^3 \) is negative, so \( -x^3/3 \) is positive and large). Wait, I made a mistake earlier: \( y = -\frac{x^3}{3}+x^2 \). As \( x \to -\infty \), \( -x^3/3 \) is positive and dominant, so \( y \to \infty \). At \( x = -1 \), \( y = 1/3 + 1 = 4/3 \). At \( x = 0 \), \( y = 0 \). So from \( x = -\infty \) to \( x = 0 \), function decreases (since \( y' < 0 \)) from \( \infty \) to \( 0 \). Then from \( x = 0 \) to \( x = 2 \), increases to \( y = 4/3 \) (at \( x = 2 \): \( y = -8/3 + 4 = 4/3 \)). Then from \( x = 2 \) to \( \infty \), decreases to \( -\infty \). So at \( x = 0 \): local minimum (since left decreasing, right increasing? Wait, no: left is decreasing (from \( \infty \) to \( 0 \)), right is increasing (from \( 0 \) to \( 4/3 \)) → so \( x = 0 \) is a local minimum. At \( x = 2 \): left increasing, right decreasing → local maximum (value \( 4/3 \)).

Step8: Sketch the graph

Plot intercepts \( (0,0) \), \( (3,0) \). Critical points: \( (0,0) \) (local min), \( (2, 4/3) \) (local max). Inflection point \( (1, y(1)) = (1, -1/3 + 1) = (1, 2/3) \). Concave up on \( (-\infty,1) \), concave down on \( (1,\infty) \). Decreasing on \( (-\infty,0) \) and \( (2,\infty) \), increasing on \( (0,2) \).

Answer:

• x-intercepts: \( (0, 0) \), \( (3, 0) \); y-intercept: \( (0, 0) \)
• Critical x-coordinates: \( 0, 2 \)
• Increasing: \( (0, 2) \); Decreasing: \( (-\infty, 0) \), \( (2, \infty) \)
• Inflection x-coordinate: \( 1 \)
• Concave up: \( (-\infty, 1) \); Concave down: \( (1, \infty) \)
• Relative min: \( x = 0 \) ( \( y = 0 \) ); Relative max: \( x = 2 \) ( \( y = \frac{4}{3} \) )
• Graph: Plot key points, use concavity and increase/decrease to sketch.