Question

Is `f(x)=(1-xe^x)/(1-x^2)` increasing or decreasing at `x=2`?

Answer 1

Function `f(x)=(1-xe^x)/(1-x^2)` is increasing at `x=2`

Explanation:

To find whether a function is increasing or decreasing at a given point say `x=x_0`, we need to differentiate it and find value of the derivative at `x=x_0`.

As `f(x)=(1-xe^x)/(1-x^2)`, using quotient formula

`(df)/(dx)=(-(1-x^2)(e^x+xe^x)+2x(1-xe^x))/(1-x^2)^2`

and at `x=2`

`(df)/(dx)=(-(1-4)(e^2+2e^2)+4(1-2e^2))/(1-4)^2`

= `(9e^2+4-8e^2)/9=(e^2+4)/9`

As `(df)/(dx)>0`, the function is increasing at `x=2`.

graph{(1-xe^x)/(1-x^2) [-9.67, 10.33, -1.12, 8.88]}

Answer 2

Increasing

Explanation:

the sign of the derivative of the function determines if its increasing or decreasing

therefore, we have to differentiate it first,
I'll use the quotient rule here

therefore derivative =

`(e^x(x^3 - x^2 - x - 1) + 2x)/(1-x^2)^2`

therefore, just put in x = 2

`(e^2(8 - 4 - 2 - 1) + 2*2)/9`

this value is positive, therefore
the function is increasing at `x = 2`