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

(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}