Q. 66 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 63–72, for the given functions f and g, find the following. For parts (a)–(d), also find the domain.

$$\begin{gathered} \left(a\right) (f+g)\left(x\right) \left(b\right) (f-g)\left(x\right) \left(c\right) (f·g) \left(x\right) \left(d\right) \left(\frac{f}{g}\right)\left(x\right) \\ \left(e\right) (f+g)\left(3\right) \left(f\right) (f-g)\left(4\right) \left(g\right) ( f·g)\left(2\right) \left(h\right) \left(\frac{f}{g}\right)\left(1\right) \end{gathered}$$

$f\left(x\right)=2x^2+3; g\left(x\right)=4x^3+1$

Step-by-Step Solution

Answer

The value of $(f+g)\left(x\right)=4x^3+2x^2+4$ and domain of $f+g$ is all set of real numbers.

The value of $(f-g)\left(x\right)=-4x^3+2x^2+2$ and the domain of $f-g$ is set of all real numbers.

The value of $(f·g)\left(x\right)=8x^5+12x^3+2x^2+3$ and the domain of $f·g$ is set of all real numbers.

The value of $\left(\frac{f}{g}\right)\left(x\right)=\frac{2x^2+3}{4x^3+1}$and the domain of $\frac{f}{g}$ is set of all real numbers.

The value of $(f+g)\left(3\right)=130$

The value of $(f-g)\left(4\right)=-238$

The value of $(f·g)\left(2\right)=363$

The value of $\left(\frac{f}{g}\right)\left(1\right)=1$

Step 1. Given Information

In the given problems we have to solve the given functions f and g, find the following. For parts (a)–(d), we have to also find the domain.

$$\begin{gathered} \left(a\right) (f+g)\left(x\right) \left(b\right) (f-g)\left(x\right) \left(c\right) (f·g) \left(x\right) \left(d\right) \left(\frac{f}{g}\right)\left(x\right) \\ \left(e\right) (f+g)\left(3\right) \left(f\right) (f-g)\left(4\right) \left(g\right) ( f·g)\left(2\right) \left(h\right) \left(\frac{f}{g}\right)\left(1\right) \end{gathered}$$

The given function is $f\left(x\right)=2x^2+3; g\left(x\right)=4x^3+1$

Step 2. The function f tells us two times of square of a number and then add three. The function g tells us four times of cube of a number and then add one.

Since these operations can be performed on any real number, we conclude that the domain of $f \mathrm{and} g$ is the set of all real numbers. 

Part (a) Step 1. We have to find the value of ( f + g ) ( x )

We know that $(f+g)\left(x\right)=f\left(x\right)+g\left(x\right)$

Part (a) Step 2. Putting the value of f ( x )   and   g ( x )

$$\begin{gathered} (f+g)\left(x\right)=2x^2+3+4x^3+1 \\ (f+g)\left(x\right)=4x^3+2x^2+4 \end{gathered}$$

The domain of $f+g$ consists of those numbers x that are in the domains of both $f \mathrm{and} g$. Therefore, the domain of $f+g$ is all set of real number.

Part (b) Step 1. We have to find the value of ( f - g ) ( x ) We know ( f - g ) ( x ) = f ( x ) - g ( x )

Putting the value of$f\left(x\right) \mathrm{and} g\left(x\right)$

$$\begin{gathered} (f-g)\left(x\right)=2x^2+3-(4x^3+1) \\ (f-g)\left(x\right)=2x^2+3-4x^3-1 \\ (f-g)\left(x\right)=-4x^3+2x^2+2 \end{gathered}$$

The domain of $f-g$ consists of those numbers x that are in the domains of both $f \mathrm{and} g$. Therefore, the domain of $f-g$ is all set of real number.

Part (c) Step 1. We have to find the value of ( f · g )   ( x ) We know that ( f · g )   ( x ) = f ( x ) · g ( x )

Putting the value of $f\left(x\right) \mathrm{and} g\left(x\right)$

$(f·g)\left(x\right)=(2x^2+3)·(4x^3+1)$

Using commutative property

$$\begin{gathered} (f·g)\left(x\right)=2x^2·(4x^3+1)+3·(4x^3+1) \\ (f·g)\left(x\right)=2x^2·4x^3+2x^2·1+3·4x^3+3·1 \\ (f·g)\left(x\right)=8x^5+12x^3+2x^2+3 \end{gathered}$$

The domain of $f·g$ consists of those numbers x that are in the domains of both $f \mathrm{and} g$. Therefore, the domain of $f·g$ is all set of real number.

Part (d) Step 1. We have to find the value of f g ( x ) We know that f g ( x ) = f ( x ) g ( x )

Putting the value of $f\left(x\right) \mathrm{and} g\left(x\right)$

$\left(\frac{f}{g}\right)\left(x\right)=\frac{2x^2+3}{4x^3+1}$

Part (d) Step 2. The domain of f g consists of the numbers for which g ( x ) ≠ 0 and that are in the domains of both f   and   g .

Since $g\left(x\right)\ne 0$ when $4x^3+1\ne 0$

The domain of $\frac{f}{g}$ is set of all real numbers.

Part (e) Step 1. We have to find the value of ( f + g ) ( 3 ) From the part (a) we know the value of ( f + g ) ( x ) = 4 x 3 + 2 x 2 + 4

Putting $x=3$ in the value of $(f+g)\left(x\right)$

$$\begin{gathered} (f+g)\left(3\right)=4{\left(3\right)}^3+2{\left(3\right)}^2+4 \\ (f+g)\left(3\right)=4·27+2·9+4 \\ (f+g)\left(3\right)=108+18+4 \\ (f+g)\left(3\right)=130 \end{gathered}$$

Part (f) Step 1. We have to find the value of ( f - g ) ( 4 ) From the part (b) we know the value of ( f - g ) ( x ) = - 4 x 3 + 2 x 2 + 2

Putting $x=4$ in the value of $(f-g)\left(x\right)$

$$\begin{gathered} (f-g)\left(4\right)=-4{\left(4\right)}^3+2{\left(4\right)}^2+2 \\ (f-g)\left(4\right)=-4·64+2·8+2 \\ (f-g)\left(4\right)=-256+16+2 \\ (f-g)\left(4\right)=-238 \end{gathered}$$

Part (g) Step 1. We have to find the value of ( f · g ) ( 2 ) From the part (c) we know the value of ( f · g ) ( x ) = 8 x 5 + 12 x 3 + 2 x 2 + 3

Putting $x=2$ in the value of $(f·g)\left(x\right)$

$$\begin{gathered} (f·g)\left(2\right)=8{\left(2\right)}^5+12{\left(2\right)}^3+2{\left(2\right)}^2+3 \\ (f·g)\left(2\right)=8·32+12·8+2·4+3 \\ (f·g)\left(2\right)=256+96+8+3 \\ (f·g)\left(2\right)=363 \end{gathered}$$

Part (h) Step 1. We have to find the value of f g ( 1 ) From the part (d) we know the value of f g ( x ) = 2 x 2 + 3 4 x 3 + 1

Putting $x=1$ in the value of $\left(\frac{f}{g}\right)\left(x\right)$

$$\begin{gathered} \left(\frac{f}{g}\right)\left(1\right)=\frac{2{\left(1\right)}^2+3}{4{\left(1\right)}^3+1} \\ \left(\frac{f}{g}\right)\left(1\right)=\frac{2·1+3}{4·1+1} \\ \left(\frac{f}{g}\right)\left(1\right)=\frac{2+3}{4+1} \\ \left(\frac{f}{g}\right)\left(1\right)=\frac{5}{5} \\ \left(\frac{f}{g}\right)\left(1\right)=1 \end{gathered}$$