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Question: If \(f(x,y) = {({\text{max(}}x,y))^{{\text{min}}(x,y)}}\) and \(g(x,y) = {\text{max}}(x,y) - \min (x...

If f(x,y)=(max(x,y))min(x,y)f(x,y) = {({\text{max(}}x,y))^{{\text{min}}(x,y)}} and g(x,y)=max(x,y)−min⁡(x,y)g(x,y) = {\text{max}}(x,y) - \min (x,y) then Value off(g(−1,−32),g(−4,−1.75))f\left( {g\left( { - 1, - \dfrac{3}{2}} \right),g( - 4, - 1.75)} \right) equals:

Explanation

Solution

If we are given that f(x)=max⁡(1,2) then f(x)=2f(x) = \max (1,2){\text{ then }}f(x) = 2 because 2 is greater than 1. Similarly here we need to find the value of (g(−1,−32),g(−4,−1.75))\left( {g\left( { - 1, - \dfrac{3}{2}} \right),g( - 4, - 1.75)} \right) and then we will find the value of f(g(−1,−32),g(−4,−1.75))f\left( {g\left( { - 1, - \dfrac{3}{2}} \right),g( - 4, - 1.75)} \right)

Complete step by step solution:
Here we are given the two functions which are f(x),g(x)f(x),g(x) and are defined as
f(x,y)=(max(x,y))min(x,y)f(x,y) = {({\text{max(}}x,y))^{{\text{min}}(x,y)}}
g(x,y)=max(x,y)−min⁡(x,y)g(x,y) = {\text{max}}(x,y) - \min (x,y)
Here max and min are given so let us understand what is the meaning of max and min so if we are given that f(x)=max⁡(x,y),x>y, then f(x)=xf(x) = \max (x,y),x > y,{\text{ then }}f(x) = x because the maximum of the two values x and yx{\text{ and }}y is xxand if we are given to find the value of min⁡(x,y) then f(x)=y\min (x,y){\text{ then }}f(x) = y because it is minimum of the given two values in the function.
Let us explain by taking the graph and here g(x) and f(x)g(x){\text{ and }}f(x) are represented in the graph as


f(x)f(x) is the curved line and g(x)g(x) is the straight line intersecting at x1,x2,x3{x_1},{x_2},{x_3}
So for the condition that x1<x<x2{x_1} < x < {x_2} we know that f(x)>g(x)f(x) > g(x) then max⁡(f(x),g(x))=f(x)\max (f(x),g(x)) = f(x) and min⁡(f(x),g(x))=g(x)\min (f(x),g(x)) = g(x)
Now for x2<x<x3{x_2} < x < {x_3} we know that g(x)>f(x)g(x) > f(x) then max⁡(f(x),g(x))=g(x)\max (f(x),g(x)) = g(x) and min⁡(f(x),g(x))=f(x)\min (f(x),g(x)) = f(x)
Now we are given that g(x,y)=max(x,y)−min⁡(x,y)g(x,y) = {\text{max}}(x,y) - \min (x,y)
We need to find the value of (g(−1,−32),g(−4,−1.75))\left( {g\left( { - 1, - \dfrac{3}{2}} \right),g( - 4, - 1.75)} \right)
So for (g(−1,−32))\left( {g\left( { - 1, - \dfrac{3}{2}} \right)} \right) =(max⁡g(−1,−32)−min⁡g(−1,−32)) = \left( {\max g\left( { - 1, - \dfrac{3}{2}} \right) - \min g\left( { - 1, - \dfrac{3}{2}} \right)} \right)
As −1>−32 - 1 > - \dfrac{3}{2} we can write that
(g(−1,−32))=−1−(−32)=12\left( {g\left( { - 1, - \dfrac{3}{2}} \right)} \right) = - 1 - ( - \dfrac{3}{2}) = \dfrac{1}{2}
Now we need to find g(−4,−1.75)g( - 4, - 1.75)
=max⁡(−4,−1.75)−min⁡(−4,−1.75) =−1.75−(−4)=2.25  = \max ( - 4, - 1.75) - \min ( - 4, - 1.75) \\\ = - 1.75 - ( - 4) = 2.25 \\\
Now we need to find the value of f(g(−1,−32),g(−4,−1.75))f\left( {g\left( { - 1, - \dfrac{3}{2}} \right),g( - 4, - 1.75)} \right)
Or we can say f(g(−1,−32),g(−4,−1.75))f\left( {g\left( { - 1, - \dfrac{3}{2}} \right),g( - 4, - 1.75)} \right) =f(0.5,2.25) = f(0.5,2.25)
Now we know that it is defined that f(x,y)=(max(x,4))min(x,4)f(x,y) = {({\text{max(}}x,4))^{{\text{min}}(x,4)}}
f(0.5,2.25)=(max(0.5,2.25))min(0.5,2.25)f(0.5,2.25) = {({\text{max(0}}{\text{.5}},2.25))^{{\text{min}}(0.5,2.25)}}=2.250.5=225100=1510=1.5{2.25^{0.5}} = \sqrt {\dfrac{{225}}{{100}}} = \dfrac{{15}}{{10}} = 1.5

So we get that f(g(−1,−32),g(−4,−1.75))f\left( {g\left( { - 1, - \dfrac{3}{2}} \right),g( - 4, - 1.75)} \right) =1.5 = 1.5

Note:
So if the function is defined like f(x)f(x) then for its being maximum and minimum the value of the derivative of it must be equal to zero which means that f′(x)=0f'(x) = 0 so now if we get its double derivative as positive then it is the minimum and if it is negative then it is the maximum.