Calc Question

Question

for what values of x is the curve f(x)=1/(X²+1) increasing and concave down?

Answer 1

Hi Shelby,

 

To determine where the function is increasing and decreasing and whether it is concave up or down we must find the first derivative of the function.  First I would write f(x) = 1/(x² + 1)  as f(x) = (x² + 1)^-1 Using the power rule and the chain rule we get

                    f'(x) = -1(x² + 1)^-2(2x) = -2x/(x² + 1)²  Set the first derivative = zero and solve for x.

 

                          -2x/(x² + 1)² = 0    Multiply both sides by the denominator resulting in  -2x = 0 and x = 0. Therefore x = 0 is a critical point. Using the first derivative test with a point on both sides of zero we get 

 

                    f'(-1) = 2/4 indicating that the function is increasing on the interval (-∞,0)

                    f'(1)  = -2/4 indicating that the function is decreasing on the interval (0,∞)

 

                    Since the function is increasing on the left of 0 and decreasing on the right side of zero, the function is concave down at x = 0

Answer 2

To find intervals of decreasing, we set the derivative of f(x) equal to zero.

 

f'(x) = 0

 

Use the quotient rule to derive.

 

-2x / (x² + 1)² = 0

 

x = 0

 

 

This is the critical point and only local extrema.  Next, we use test points and evaluate the derivative using them.  Evaluate f'(-1) and f'(1).

 

f'(-1) = (-2 * -1) / [(-1)² + 1)]²

         = 2 / 4

         

 

f'(1) = (-2 * 1) / [1² + 1]²

       = -2 / 4

 

 

Since the derivative is positive at x=1, f(x) increases  when  x > 0.

 

 

 

To find where it is concave down, set the second derivative of f(x) equal to zero.

 

[-2(x² + 1)² - 2(x² + 1)(2x)(-2x)] / [(x² + 1)⁴] = 0

 

[-2(x⁴ + 2x² + 1) + (8x⁴ + 8x²)] / (x² + 1)⁴ = 0

 

[-2x⁴ - 4x² - 2 + 8x⁴ + 8x²] / (x² + 1)⁴ = 0

 

(6x⁴ + 4x² - 2)  / (x² + 1)⁴ = 0

 

 

[2(3x⁴ + 2x² - 1 )] / (x² + 1)⁴ = 0

 

 

3x⁴ + 2x² - 1 = 0

 

(3x² - 1)(x² + 1) = 0

 

 

3x² - 1 = 0

 

3x² = 1

 

x² = ±1 / √3

 

x = -√(3) / 3             and            x = √(3) / 3

 

 

The x values are your points of inflections  Next, use test points around the points of inflection.  Use those test points to evaluate the second derivative.  Since you want concave down, you want to look for a negative second derivative.