Chapter 15, Problem 1RE

To determine

To calculate: The value of $f\left(0,1,1\right),f\left(2,1,1\right),f\left(-1,1,-1\right),f\left(z,z,z\right)$ and $f\left(x+h,y+k,z+l\right)$ if $f\left(x,y,z\right)=\frac{x}{y+xz}+x^{2}y$.

Answer to Problem 1RE

Solution:

For the provided function $f\left(0,1,1\right)=0$, $f\left(2,1,1\right)=\frac{14}{3}$, $f\left(-1,1,-1\right)=\frac{1}{2}$, $f\left(z,z,z\right)=\frac{1}{1+z}+z^3$ and $f\left(x+h,y+k,z+l\right)=\frac{x+h}{y+k+\left(x+h\right)\left(z+l\right)}+{\left(x+h\right)}^2\left(y+k\right)$.

Explanation of Solution

Given Information:

The provided function is,

$f\left(x,y,z\right)=\frac{x}{y+xz}+x^{2}y$

Formula used:

A real-valued function, $f$, of $x,y,z$,… is a rule for manufacturing a new number, written $f\left(x,y,z,\mathrm{...}\right)$ from the values of a sequence of independent variables $\left(x,y,z,\dots \right)$.

Calculation:

Consider the provided function,

$f\left(x,y,z\right)=\frac{x}{y+xz}+x^{2}y$

Now, to evaluate $f\left(0,1,1\right)$, substitute $0$ for $x$, $1$ for $y$ and $1$ for $z$ in the above function to get,

$\begin{array}{c}f\left(0,1,1\right)=\frac{0}{1+0\left(1\right)}+0^2\left(1\right)\\ =0\end{array}$

Thus, $f\left(0,1,1\right)=0$

To evaluate $f\left(2,1,1\right)$, substitute $2$ for $x$, $1$ for $y$ and $1$ for $z$ in the above function to get,

$\begin{array}{c}f\left(2,1,1\right)=\frac{2}{1+2\left(1\right)}+2^2\left(1\right)\\ =\frac{2}{3}+4\\ =\frac{14}{3}\end{array}$

Thus, $f\left(2,1,1\right)=\frac{14}{3}$.

To evaluate $f\left(-1,1,-1\right)$, substitute $-1$ for $x$, $1$ for $y$ and $-1$ for $z$ in the above function to get,

$\begin{array}{c}f\left(-1,1,-1\right)=\frac{-1}{1+\left(-1\right)\left(-1\right)}+{\left(-1\right)}^2\left(1\right)\\ =-\frac{1}{2}+1\\ =\frac{1}{2}\end{array}$

Thus, $f\left(-1,1,-1\right)=\frac{1}{2}$.

To evaluate $f\left(z,z,z\right)$, substitute $z$ for $x$, $z$ for $y$ and $z$ for $z$ in the above function to get,

$\begin{array}{c}f\left(z,z,z\right)=\frac{z}{z+z\left(z\right)}+z^2\left(z\right)\\ =\frac{z}{z+z^2}+z^3\\ =\frac{1}{1+z}+z^3\end{array}$

Thus, $f\left(z,z,z\right)=\frac{1}{1+z}+z^3$.

To evaluate $f\left(x+h,y+k,z+l\right)$, substitute $x+h$ for $x$, $y+k$ for $y$ and $z+l$ for $z$ in the above function to get,

$f\left(z+h,y+,z+l\right)=\frac{x+h}{y+k+\left(x+h\right)\left(z+l\right)}+{\left(x+h\right)}^2\left(y+k\right)$

Thus, $f\left(x+h,y+k,z+l\right)=\frac{x+h}{y+k+\left(x+h\right)\left(z+l\right)}+{\left(x+h\right)}^2\left(y+k\right)$.

Thus, for the provided function $f\left(0,1,1\right)=0$, $f\left(2,1,1\right)=\frac{14}{3}$, $f\left(-1,1,-1\right)=\frac{1}{2}$, $f\left(z,z,z\right)=\frac{1}{1+z}+z^3$ and $f\left(x+h,y+k,z+l\right)=\frac{x+h}{y+k+\left(x+h\right)\left(z+l\right)}+{\left(x+h\right)}^2\left(y+k\right)$.