Is f(x)=(1-xe^x)/(1-x^2)f(x)=1−xex1−x2 increasing or decreasing at x=2x=2?

2 Answers
Shwetank Mauria
Mar 9, 2018

Function f(x)=(1-xe^x)/(1-x^2)f(x)=1−xex1−x2 is increasing at x=2x=2

Explanation:

To find whether a function is increasing or decreasing at a given point say x=x_0x=x0, we need to differentiate it and find value of the derivative at x=x_0x=x0.

As f(x)=(1-xe^x)/(1-x^2)f(x)=1−xex1−x2, using quotient formula

(df)/(dx)=(-(1-x^2)(e^x+xe^x)+2x(1-xe^x))/(1-x^2)^2dfdx=−(1−x2)(ex+xex)+2x(1−xex)(1−x2)2

and at x=2x=2

(df)/(dx)=(-(1-4)(e^2+2e^2)+4(1-2e^2))/(1-4)^2dfdx=−(1−4)(e2+2e2)+4(1−2e2)(1−4)2

= (9e^2+4-8e^2)/9=(e^2+4)/99e2+4−8e29=e2+49

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

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

dhruv ahlawat
Mar 9, 2018

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)^2ex(x3−x2−x−1)+2x(1−x2)2

therefore, just put in x = 2

(e^2(8 - 4 - 2 - 1) + 2*2)/9e2(8−4−2−1)+2⋅29

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

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