Question

The functions in Problems 1 to 3 are neither even nor odd. Write each of them as the sum of an even function and an odd function.

(a) ${\mathit{x}}^{\mathbf{5}}\mathbf{-}{\mathit{x}}^{\mathbf{4}}\mathbf{+}{\mathit{x}}^{\mathbf{3}}\mathbf{-}\mathbf{1}$

(b)$\mathbf{1}\mathbf{+}{\mathit{e}}^{\mathbf{x}}$

Short answer

Answer

The functions in the form of the sum of an even function $\left(f_e\left(x\right)\right)$and an odd function $\left(f_e\left(x\right)\right)$for $x^5-x^4+x^3-1$are $-x^4-1$and $x^5+x^3$respectively. Similarly, for $1+e^x$, the functions are $1+\cos hx$and role="math" localid="1654940194928" $\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathit{x}$respectively.

Step-by-step solution

Given Information.

The two given functions are $x^5-x^4+x^3-1$and$1+e^x$.

The significance of Even and Odd Functions.

A function $\mathit{f}\left(x\right)$is an even function when$\mathit{f}\left(-x\right)\mathbf{=}\mathit{f}\left(x\right)$, where as a function $\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$is an odd function when$\mathit{f}\mathbf{(}\mathbf{-}\mathbf{x}\mathbf{)}\mathbf{=}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.

Write the functions as the sum of an even function and odd function.

The two functions are $x^5-x^4+x^3-1$and$1+e^x$.

(a) $x^5-x^4+x^3-1$

Express the function $x^5-x^4+x^3-1$as the sum of an even function.

$$\begin{gathered} \left(f_e\left(x\right)\right)=\frac{1}{2}\left(f\left(x\right)+f\left(-x\right)\right) \\ \left(f_0\left(x\right)\right)=\left(\frac{x^5-x^4+x^3-1+\left(-x^5-x^4-x^3-1\right)}{2}\right) \end{gathered}$$

Simplify further as shown below.

$$\begin{gathered} \left(f_0\left(x\right)\right)=\left(\frac{x^5-x^4+x^3-1+\left(-x^5-x^4-x^3-1\right)}{2}\right) \\ \left(f_e\left(x\right)\right)=-x^4-1 \end{gathered}$$

Express the function $x^5-x^4+x^3-1$as the sum of an odd function.

$$\begin{gathered} \left(f_e\left(x\right)\right)=\frac{1}{2}\left(f\left(x\right)+f\left(-x\right)\right) \\ \left(f_0\left(x\right)\right)=\left(\frac{x^5-x^4+x^3-1+\left(-x^5-x^4-x^3-1\right)}{2}\right) \end{gathered}$$

Simplify further as shown below.

$\left(f_e\left(x\right)\right)=x^5+x^3$

The sum of the even and odd function is expressed as:

$$\begin{gathered} f\left(x\right)=f_e\left(x\right)+f_0\left(x\right) \\ =-x^{}-1+x^5+x^3 \\ =x^5-x^4+x^3-1 \end{gathered}$$

Hence, the sum of the even and odd function is, $x^5-x^4+x^3-1$.

(b) $1+e^x$

Express the function $1+e^x$as the sum of an even function.

$$\begin{gathered} \left(f_0\left(x\right)\right)=\frac{1}{2}\left(f\left(x\right)-f\left(-x\right)\right) \\ \left(f_0\left(x\right)\right)=\left(\frac{1+e^x-1-e^x}{2}\right) \end{gathered}$$

Simplify further as shown below.

$\left(f_e\left(x\right)\right)=1+\cos hx$

Express the function$1+e^x$as the sum of an odd function.

$$\begin{gathered} f\left(x\right)=f_0\left(x\right)+f_0\left(x\right) \\ =1+\cos hx+\sin hx \end{gathered}$$

Simplify further as shown below.

$\left(f_0\left(x\right)\right)=\sin hx$

The sum of the even and odd function is expressed as:

$$\begin{gathered} f\left(x\right)=f_0\left(x\right)+f_0\left(x\right) \\ =1+\cos hx+\sin hx \end{gathered}$$

Hence, the sum of the even and odd function is, role="math" localid="1654941711471" $1+\cos hx+\sin hx$.

Therefore, the function$x^5-x^4+x^3-1$can be expressed in the form of the sum of an even function$\left(f_e\left(x\right)\right)$and an odd function $\left(f_0\left(x\right)\right)$as $-x^4-1$and $x^5+x^3$respectively. Similarly, the function $1+e^x$can be expressed in the form of the sum of an even function $\left(f_e\left(x\right)\right)$and an odd function $\left(f_0\left(x\right)\right)$as $1+\cos hx$and $\sin hx$respectively.