two questions?

Question

1. [2e^(-x) - xe^(-x)] / (1-x^2) > 0

2. (4x + 1)^(1/3) + (4/3)x (4x + 1)^(-2/3) = 0

your assistance would be greatly apprieciated and the winner or right answer will be awarded my ten points. thanks and good luck!!

Answer 1

1. I think you want to find the values x where this function is positive, correct?

You have f(x) = e^(-x)(2 - x)/(1 - x)(1 + x). Since the exponential e^(-x) is always positive, the sign of the function is determined completely by F(x) = (2 - x)/(x - 1)(x + 1). Notice that the numerator is = 0 at x = 2, and the denominator = 0 at x = -1 and x = 1. The function can only change sign at those three values, -1, 1, and 2. These numbers divide the line into four regions; any number within those regions will tell us of the sign of F(x) THROUGHOUT THE ENTIRE REGION. We may choose any convenient number in each region. Let us take x = -2, 0, 3/2, 3. We find F(-2) > 0,
F(0) < 0, F(3/2) > 0, and F(3) > 0.

Summarizing results, we know that F(x) (and hence f(x)) is positive for x in (-inf,-1), or in (1,2) or in (2,inf).

2. Multiply this equation by (4x + 1)^(2/3), to get
(4x + 1) + (4/3)x = 0. Solve for x.

Answer 2

first problem=e^(-x)(2-x)/(1-x)(1+x)

2-x>0, x<2

(1-x)(1+x)>0 and

-1
1/e^x>0

-1
2=(4x+1)^1/3[1+4/3x(4x+1)^-1/3]=0

(4x+1)^1/3=0,x=-1/4,

Answer 3

I don't understand !!!!!!!!!