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

Question

Elijah Abram · Accepted Answer

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

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

Christopher Allers · Answer

Increasing

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#

Charles Anctil · Answer

undefined To determine if the function $ f(x) = \frac{1 - xe^x}{1 - x^2} $ is increasing or decreasing at $ x = 2 $, we evaluate the derivative of the function at that point. The derivative of $ f(x) $ is given by:

$$ f'(x) = \frac{(1 - x^2)(-e^x) - (1 - xe^x)(-2x)}{(1 - x^2)^2} $$

Now, we substitute $ x = 2 $ into the derivative:

$$ f'(2) = \frac{(1 - 2^2)(-e^2) - (1 - 2e^2)(-2(2))}{(1 - 2^2)^2} $$

Simplifying this expression gives us:

$$ f'(2) = \frac{(1 - 4)(-e^2) - (1 - 2e^2)(-4)}{(1 - 4)^2} $$
$$ f'(2) = \frac{3e^2 + 4(1 - 2e^2)}{9} $$
$$ f'(2) = \frac{12 - 5e^2}{9} $$

Now, to determine if $ f(x) $ is increasing or decreasing at $ x = 2 $, we look at the sign of $ f'(2) $. Since $ f'(2) $ is negative (given that $ e^2 $ is approximately 7.389), we can conclude that the function $ f(x) $ is decreasing at $ x = 2 $.