Prove that (x-1)*f(x)>0 if f(x)=(x^4+x)/x(x^2-1)

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have the function f(x) = (x^4 + x)/x*(x^2 - 1)

(x - 1)*f(x) = (x - 1)(x^4 + x)/x*(x^2 - 1)

=> (x - 1)*x*(x^3 + 1)/x*(x - 1)(x + 1)

=> (x^3 + 1)/(1 + x)

=> (1 + x)(x^2 - x + 1)/(1 + x)

=> (x^2 - x + 1)

=> x^2 - x + 1/4 + 3/4

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

Here (x - 1/2)^2 is positive and 3/4 is positive. The sum of two positive numbers is positive.

This proves that (x-1)*f(x) > 0

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We need to factorize the numerator of the function:

f(x) = x(x^3 + 1)/x(x^2 - 1)

We'll simplify:

f(x) = (x^3 + 1)/(x^2 - 1)

The sum of cubes form numerator returns the product:

x^3+1 = (x+1)(x^2 - x + 1)

The difference of squares form numerator returns the product:

x^2 - 1 = (x-1)(x+1)

f(x) = (x+1)(x^2 - x + 1)/(x-1)(x+1)

We'll simplify:

f(x) = (x^2 - x + 1)/(x-1)

Now, we have to verify if (x-1)*f(x)>0

(x -1)*(x^2 - x + 1)/(x-1) > 0

We'll simplify:

(x^2 - x + 1) > 0

We'll verify if the parabola given by the quadratic expression is above x axis.

For this reason, we'll check the value of it's discriminant:

delta = b^2 - 4ac

a = 1, b = -1 , c = 1

delta = 1 - 4 = -3 < 0

Since delta is negative and a = 1>0, the parabola given by the quadratic expression is above x axis.

Therefore, the inequality (x-1)*f(x)>0 is verified.

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