Answer:
The interval for which y is a decreasing function of x is:
[tex](0, 1)[/tex]
Or as an inequality:
[tex]0<x<1[/tex]
Step-by-step explanation:
We are given the equation:
[tex]\displaystyle y=\frac{e^x}{x}\, ,x>0[/tex]
And we want to find the range of values x for which y is a decreasing function of x.
y is decreasing whenever y' is negative. Find y' using the Quotient Rule:
[tex]\displaystyle y'=\frac{(e^x)'(x)-e^x(x)'}{(x)^2}[/tex]
Differentiate:
[tex]\displaystyle y'=\frac{xe^x-e^x}{x^2}[/tex]
y is decreasing whenever y' is negative. Thus:
[tex]\displaystyle 0>\frac{xe^x-e^x}{x^2}[/tex]
Multiply both sides by x². This is always positive so we do not need to change the sign:
[tex]xe^x-e^x<0[/tex]
Factor:
[tex]e^x(x-1)<0[/tex]
eˣ is always positive. So:
[tex]x-1<0[/tex]
Adding one to both sides produces:
[tex]x<1[/tex]
Therefore, y is a decreasing function of x when x is less than one (and greater than 0).
In interval notation:
[tex](0, 1)[/tex]
Or as an inequality:
[tex]0<x<1[/tex]
