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Graph each function using a graphing calculator by first making a sign diagram for just first...

Question:

Graph each function using a graphing calculator by first making a sign diagram for just first derivative. Make a sketch from the screen, showing the coordinates of all relative extreme points and inflection point. Graph may vary depending on the window chosen. {eq}1.30x^\frac{1}{3-10x}\\ 2.f(x)=6x-10x^\frac{3}{5} \\3.f(x)=3x^\frac{4}{3-2x^2} {/eq} For each function find all critical number and then use the second-derivative test to determine whether the function has a relative maximum or minimum at each critical number. {eq}1.f(x)= x^3-12x+4 {/eq}

Critical points:

With the first derivative of the function, we can find the possible maximum and minimum points of the function, we can also graph the function, with the graph we can easily observe the critical points.

Answer and Explanation:

Part 1.

1. We have,

{eq}f(x)=30x^\frac{1}{3-10x}\\ {/eq}

Differentiating the function

{eq}f'(x)=30\,{x}^{ \left( 3-10\,x \right) ^{-1}} \left( 10\,{\frac {\ln \left( x \right) }{ \left( 3-10\,x \right) ^{2}}}+{\frac {1}{ \left( 3-10\,x \right) x}} \right) \\ {/eq}

{eq}f'(x)=DNE {/eq} when

{eq}x=0 {/eq} and {eq}f'(x)=0 {/eq} when {eq}x=0.08 \\ {/eq}

Differentiating the function

{eq}f''(x)=60\,{\frac {50\, \left( \ln \left( x \right) \right) ^{2}{x}^{2}- 1000\,\ln \left( x \right) {x}^{3}+200\,\ln \left( x \right) {x}^{2} +1500\,{x}^{3}+30\,\ln \left( x \right) x-1000\,{x}^{2}+195\,x-9}{ \left( -3+10\,x \right) ^{4}}{x}^{-5\,{\frac {-1+4\,x}{-3+10\,x}}}} {/eq}

{eq}f''(x)=0 {/eq} The function has no real solution, therefore, the function has no inflection point.

Graph:

Relative minimum at:

{eq}(0,0) \\ {/eq}

Relative maximum at:

{eq}(0.08, 9.51) \\ {/eq}

2. We have

{eq}f(x)=6x-10x^\frac{3}{5} \\ {/eq}

Differentiating the function

{eq}f'(x)=6-6\,{x}^{-2/5} {/eq}

{eq}f'(x)=0 {/eq} when {eq}x=1 {/eq} and {eq}f'(x)=DNE {/eq} when {eq}x=0 {/eq}

Graph the function

{eq}f(x)=6x-10x^\frac{3}{5} \\ {/eq}

Relative minimum at:

{eq}(1, -4) \\ {/eq}

Now,

Differentiating the function

{eq}f''(x)={\frac {12}{5\,{x}^{7/5}}} \\ {/eq}

{eq}f''(x)=0 {/eq} The function has no real solution, therefore, the function has no inflection point.

3. We have,

{eq}f(x)=3x^\frac{4}{3-2x^2} \\ {/eq}

Differentiating the function

{eq}f'(x)=3\,{x}^{4\, \left( -2\,{x}^{2}+3 \right) ^{-1}} \left( 16\,{\frac { \ln \left( x \right) x}{ \left( -2\,{x}^{2}+3 \right) ^{2}}}+4\,{ \frac {1}{ \left( -2\,{x}^{2}+3 \right) x}} \right) {/eq}

{eq}f'(x)=DNE {/eq} when {eq}x=0 {/eq}

Differentiating the function

{eq}f''(x)=12\,{\frac {-48\,\ln \left( x \right) {x}^{6}+64\, \left( \ln \left( x \right) \right) ^{2}{x}^{4}+40\,{x}^{6}-16\,\ln \left( x \right) {x}^{4}-116\,{x}^{4}+132\,\ln \left( x \right) {x}^{2}+78\,{ x}^{2}+9}{ \left( 2\,{x}^{2}-3 \right) ^{4}}{x}^{-2\,{\frac {2\,{x}^{2 }-1}{2\,{x}^{2}-3}}}} \\ {/eq}

{eq}f''(x)=0 {/eq} The function has no real solution, therefore, the function has no inflection point.

Graph the function:

Relative minimum at: {eq}(0,0) {/eq}

Part 2.

We have:

{eq}f(x)= x^3-12x+4 \\ {/eq}

Differentiating the function

{eq}f'(x)=3\,{x}^{2}-12 \\ {/eq}

{eq}3\,{x}^{2}-12 =0 {/eq}

{eq}\begin{array}{l} \begin{array}{l} \text{This is a}\space{}{\text{second degree}}\space{}\text{equation with the form:}\space{}a x ^2+b x + c=0,\space{}\text{where}\space{}a= 3 ,\space{}b= 0 \space{}\text{and}\space{}c= -12 .\space{}\text{The solution is:} \end{array} \\ \begin{array}{lcl} x &=&\frac{-b\pm\sqrt{b^2-4*a*c}}{2*a} \\ &=&\frac{ - ( 0 ) \pm\sqrt{ ( 0 )^2 - 4 ( 3 ) ( -12 )}}{ 2 ( 3 ) } \\ &=&\frac{ 0 \pm\sqrt{ 0 - ( -144 ) }} { 6 } \\ &=&\frac{ 0 \pm\sqrt{ 0 + 144 }} { 6 } \\ &=&\frac{ 0 \pm\sqrt{ 144 }} { 6 } \\ &=&\frac{ 0 \pm 12 } { 6 } \end{array} \\ \text{First solution:} \\ \begin{array}{lcl} x _1&=&\frac{ 0 + 12 } { 6 } \\ &=&\frac{ 12 } { 6 } \\ &=& 2 \end{array} \\ \text{Second solution:} \\ \begin{array}{lcl} x _2&=&\frac{ 0 - 12 } { 6 } \\ &=&\frac{ -12 } { 6 } \\ &=& -2 \end{array} \end{array} {/eq}

Differentiating the function

{eq}f''(x)=6\,x \\ f''(2)=12 \\ f''(-2)=-12 \\ {/eq}

Relative maximum at:

{eq}(-2, 20) {/eq}

Relative minimum at:

{eq}(2, -12) \\ {/eq}

Graph the function


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Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9
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