Find each of the following functions and state their domains. (Enter the domains in interval...

Question:

Find each of the following functions and state their domains. (Enter the domains in interval notation.) {eq}f(x)=x^3+5x^2 {/eq}, {eq}g(x)=6x^2-1 {/eq}

(a) f + g

(b) f - g

(c) fg

(d) f/g

Function in One Variable:

There are two types of functions in real analysis. Function in single variable and function in multivariable. The function is one variable contains only one independent variable while multivariable function consists of more than {eq}1 {/eq} independent variables. For example, {eq}f(x)=x^2+1 {/eq} is a function in one variable while {eq}f(x,y,z)=x^2+1+y+z^3 {/eq} is a multivariable function.

Answer and Explanation: 1

(a)

Apply the sum rule {eq}(f+g)(x)=f(x)+g(x) {/eq}, where {eq}f(x)=x^3+5x^2 {/eq} and {eq}g(x)=6x^2-1 {/eq}.

{eq}\begin{aligned} \left( f+g \right)\left( x \right)&={{x}^{3}}+5{{x}^{2}}+6{{x}^{2}}-1 \\ &={{x}^{3}}+11{{x}^{2}}-1 \end{aligned} {/eq}

So, {eq}\boxed{\left( \mathbf{f+g} \right)\left( \mathbf{x} \right)\mathbf{=}{{\mathbf{x}}^{\mathbf{3}}}\mathbf{+11}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-1}} {/eq}.

Since the function {eq}\left( f+g \right)\left( x \right)={{x}^{3}}+11{{x}^{2}}-1 {/eq} is defined for all real numbers, the domain is {eq}\boxed{\left( -\infty ,\infty \right)} {/eq}.

(b)

Apply the difference rule {eq}(f-g)(x)=f(x)-g(x) {/eq}, where {eq}f(x)=x^3+5x^2 {/eq} and {eq}g(x)=6x^2-1 {/eq}.

{eq}\begin{aligned} \left( f-g \right)\left( x \right)&=\left( {{x}^{3}}+5{{x}^{2}} \right)-\left( 6{{x}^{2}}-1 \right) \\ &={{x}^{3}}+5{{x}^{2}}-6{{x}^{2}}+1 \\ &={{x}^{3}}-{{x}^{2}}+1 \end{aligned} {/eq}

So, {eq}\boxed{\left( \mathbf{f-g} \right)\left( \mathbf{x} \right)\mathbf{=}{{\mathbf{x}}^{\mathbf{3}}}\mathbf{-}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+1}} {/eq}.

Since the function {eq}\left( f-g \right)\left( x \right)={{x}^{3}}-{{x}^{2}}+1 {/eq} is defined for all real numbers, the domain is {eq}\boxed{\left( -\infty ,\infty \right)} {/eq}.

(c)

Apply the product rule {eq}(f \cdot g)(x)=f(x) \cdot g(x) {/eq}, where {eq}f(x)=x^3+5x^2 {/eq} and {eq}g(x)=6x^2-1 {/eq}.

{eq}\begin{aligned} \left( fg \right)\left( x \right)&=\left( {{x}^{3}}+5{{x}^{2}} \right)\cdot \left( 6{{x}^{2}}-1 \right) \\ &={{x}^{3}}\left( 6{{x}^{2}} \right)+{{x}^{3}}\left( -1 \right)+5{{x}^{2}}\left( 6{{x}^{2}} \right)+5{{x}^{2}}\left( -1 \right) \\ &=6{{x}^{5}}-{{x}^{3}}+30{{x}^{4}}-5{{x}^{2}} \\ &=6{{x}^{5}}+30{{x}^{4}}-{{x}^{3}}-5{{x}^{2}} \end{aligned} {/eq}

So, {eq}\left( \mathbf{fg} \right)\left( \mathbf{x} \right)\mathbf{=6}{{\mathbf{x}}^{\mathbf{5}}}\mathbf{+30}{{\mathbf{x}}^{\mathbf{4}}}\mathbf{-}{{\mathbf{x}}^{\mathbf{3}}}\mathbf{-5}{{\mathbf{x}}^{\mathbf{2}}} {/eq}.

Since the function {eq}\left( fg \right)\left( x \right)=6{{x}^{5}}+30{{x}^{4}}-{{x}^{3}}-5{{x}^{2}} {/eq} is defined for all real numbers, the domain is {eq}\boxed{\left( -\infty ,\infty \right)} {/eq}.

(d)

Apply the quotient rule {eq}\left( \dfrac{f}{g} \right)\left( x \right)=\dfrac{f(x)}{g(x)} {/eq} such that {eq}g(x) \ne 0 {/eq}, where {eq}f(x)=x^3+5x^2 {/eq} and {eq}g(x)=6x^2-1 {/eq}.

{eq}\left( \dfrac{f}{g} \right)\left( x \right)=\dfrac{{{x}^{3}}+5{{x}^{2}}}{6{{x}^{2}}-1} {/eq}

So, {eq}\boxed{\left( \dfrac{\mathbf{f}}{\mathbf{g}} \right)\left( \mathbf{x} \right)\mathbf{=}\dfrac{{{\mathbf{x}}^{\mathbf{3}}}\mathbf{+5}{{\mathbf{x}}^{\mathbf{2}}}}{\mathbf{6}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-1}}} {/eq}.

The rational function {eq}\left( \dfrac{f}{g} \right)\left( x \right)=\dfrac{{{x}^{3}}+5{{x}^{2}}}{6{{x}^{2}}-1} {/eq} is defined for all real numbers except for {eq}6x^2-1=0 {/eq}.

Solve {eq}6x^2-1=0 {/eq} for {eq}x {/eq}.

{eq}\begin{aligned} 6{{x}^{2}}-1&=0 \\ 6{{x}^{2}}&=1 \\ {{x}^{2}}&=\frac{1}{6} \\ x&=\pm \frac{\sqrt{6}}{6} \end{aligned} {/eq}

Exclude the values {eq}x=\pm \dfrac{\sqrt{6}}{6} {/eq} from the set of real number to determine the domain for {eq}\left( \dfrac{f}{g} \right)\left( x \right)=\dfrac{{{x}^{3}}+5{{x}^{2}}}{6{{x}^{2}}-1} {/eq}.

{eq}\boxed{\left( -\infty ,-\dfrac{\sqrt{6}}{6} \right)\cup \left( -\dfrac{\sqrt{6}}{6},\dfrac{\sqrt{6}}{6} \right)\cup \left( \dfrac{\sqrt{6}}{6},\infty \right)} {/eq}


Learn more about this topic:

Loading...
Algebraic Function: Definition & Examples

from

Chapter 16 / Lesson 12
73K

Learn about different types of algebraic functions and how to identify them. Also, learn the different identities of algebraic functions and see examples.


Related to this Question

Explore our homework questions and answers library