Question

Consider the multivariable function f\left(x,y,z\right)=\frac{x^3-2.5y^2+\frac{z}{2}}{z\left(x-y\right)}f ( x , y , z ) = x 3 − 2.5 y 2 + z 2 z ( x − y ). Evaluate f\left(2,3,1\right)f ( 2 , 3 , 1 ). Round your answer to two decimal places.

Answer 1

Answer:

$f\left(2,3,1\right)=\bold{14}$

Step-by-step explanation:

Given the function:

$f\left(x,y,z\right)=\frac{x^3-2.5y^2+\frac{z}{2}}{z\left(x-y\right)}$

To find:

The value of $f(2, 3, 1)$ = ?

Solution:

$f(2, 3, 1)$ means the values of $x, y\ and\ z$ as:

$x=2\\y=3\\z=1$

Let us put the given values in the given function and let us solve for it:

$\Rightarrow f\left(2,3,1\right)=\dfrac{2^3-2.5\times 3^2+\frac{1}{2}}{1\left(2-3\right)}\\\\\Rightarrow f\left(2,3,1\right)=\dfrac{8-2.5\times 9+0.5}{1\left(-1\right)}\\\\\Rightarrow f\left(2,3,1\right)=\dfrac{8-22.5+0.5}{-1}\\\\\Rightarrow f\left(2,3,1\right)=\dfrac{-14}{-1}\\\\\Rightarrow f\left(2,3,1\right)=\bold{14}$

Therefore, the answer is:

$f\left(2,3,1\right)=\bold{14}$