Use a sign chart to determine the intervals on which the function f(x)= 4x^3 - x^4 is concave up...

Question:

Use a sign chart to determine the intervals on which the function {eq}f(x)= 4x^3 - x^4 {/eq} is concave up and concave down, and identify the locations of any inflection points.

Concavity and the Second Derivative:

We say that a function {eq}f(x) {/eq} is concave up on an interval {eq}I {/eq} if the graph of {eq}y=f(x) {/eq} lies below the line segment from {eq}(a,f(a)) {/eq} to {eq}(b,f(b)) {/eq} for any {eq}a {/eq} and {eq}b {/eq} in {eq}I {/eq}. If {eq}f(x) {/eq} is continuous on the closed interval {eq}[a,b] {/eq} and twice-differentiable on the open interval {eq}(a,b) {/eq}, and {eq}f''(x)>0 {/eq} on the open interval {eq}(a,b) {/eq}, then {eq}f(x) {/eq} is concave up on the closed interval {eq}[a,b] {/eq}.

Similarly, {eq}f(x) {/eq} is concave down on an interval {eq}I {/eq} if the graph of {eq}y=f(x) {/eq} lies above the line segment from {eq}(a,f(a)) {/eq} to {eq}(b,f(b)) {/eq} for any {eq}a {/eq} and {eq}b {/eq} in {eq}I {/eq}. If {eq}f(x) {/eq} is continuous on the closed interval {eq}[a,b] {/eq} and twice-differentiable on the open interval {eq}(a,b) {/eq}, and {eq}f''(x)<0 {/eq} on the open interval {eq}(a,b) {/eq}, then {eq}f(x) {/eq} is concave down on the closed interval {eq}[a,b] {/eq}.

A point {eq}x=c {/eq} is an inflection point of {eq}f(x) {/eq} if {eq}f(x) {/eq} switches concavity at {eq}x=c {/eq}: that is, if {eq}f(x) {/eq} is concave up on some interval {eq}(a,c] {/eq} and concave down on some interval {eq}[c,b) {/eq}, or vice versa. If {eq}f''(x) {/eq} changes sign at {eq}x=c {/eq}, then {eq}x=c {/eq} is an inflection point of {eq}f(x) {/eq}.

Answer and Explanation:

Differentiating {eq}f(x)=4x^3-x^4 {/eq} twice, we have:

{eq}\begin{align*} f'(x)&=12x^2-4x^3\\ f''(x)&=24x-12x^2\\ &=12x(2-x) \, . \end{align*} {/eq}

The table below gives the sign of each of the factors of {eq}f''(x) {/eq}, and of {eq}f''(x) {/eq} itself, on every interval where that sign is constant:

{eq}\begin{array}{c|ccc} \text{range of }x&12x&2-x&f''(x)\\ \hline x < 0 & - & + & -(+)=-\\ 0 < x < 2 & + & + & +(+)=+\\ 2 < x & + & - & +(-)=- \end{array} {/eq}

So {eq}f''(x)<0 {/eq} for {eq}x<0 {/eq} and {eq}2<x {/eq}, which means that {eq}f(x) {/eq} is concave down on the intervals {eq}\left(-\infty,0\right] {/eq} and {eq}\left[2,\infty\right) {/eq}. Similarly, {eq}f''(x)>0 {/eq} for {eq}0 < x < 2 {/eq}, which means that {eq}f(x) {/eq} is concave up on the interval {eq}[0,2] {/eq}.

Because {eq}f(x) {/eq} changes concavity at {eq}x=0 {/eq} and {eq}x=2 {/eq}, {eq}f(x) {/eq} has inflection points at {eq}x=0 {/eq} and {eq}x=2 {/eq}.


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Concavity and Inflection Points on Graphs

from Math 104: Calculus

Chapter 9 / Lesson 5
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