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

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

Explanation:

Given the function:

`f\left(x,y,z\right)=(x^3-2.5y^2+(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)=(2^3-2.5* 3^2+(1)/(2))/(1\left(2-3\right))\\\\\Rightarrow f\left(2,3,1\right)=(8-2.5* 9+0.5)/(1\left(-1\right))\\\\\Rightarrow f\left(2,3,1\right)=(8-22.5+0.5)/(-1)\\\\\Rightarrow f\left(2,3,1\right)=(-14)/(-1)\\\\\Rightarrow f\left(2,3,1\right)=\bold{14}`

Therefore, the answer is:

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