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Calculus/need help on curve sketching

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Expert: Scotto - 6/25/2009


Question
Given the following function, state the domain, all asymptotes, intercepts, relative extrema and inflection points.

h(x)=(4-x^2)/(x^2+1)

I know I have to use 2nd derivative to find the inflection point but I am stuck. Need help asap :D

thanks

Answer
The function is f(x) = (4 - x²)/(x² + 1).

To find the derivative, use the quotient rule.
That is, (lo d hi - hi d low)/lo sqaured.

The first derivative is f'(x) = ((x²+1)(-2x) - (4-x²)(2x))/(x²+1)².
Simplifying this gives (-2x³ - 2x - 8x + 2x³)/(x² + 1)².

Noticing the cubes in the numerator cancel, we are left with
-10x/(x² + 1)².

Use the quotient rule again to find f"(x).
That is, f"(x) = ((x²+1)²(-10) -- 10x(2(x²+1)2x))/(x²+1)^4.

Simplifying again gives us
f"(x) = (-10(x^4 + 2x² + 1) + 10x(4x^3 + 4x))/(x²+1)^4, which is
= 10(-x^4 + 4x^3 - 2x² + 4x - 1)/(x²+1)^4.

Where this function is 0 is where the points of inflection are at.
To find these points, set -x^4 + 4x^3 - 2x² + 4x - 1 = 0 and solve.

A good way to do this would be to use a spreadsheet and put the x values in column A and then the y values in column B.
Say A1 contains an X value.  
Then B1 would have =-1+A1*(4+A1*(-2+A1*(4-A1))).

Copy B1 down to several rows and try various values column A.

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