SOLUTION: Find the intervals of increase and decrease. Use the first derivatives test to find the local maxima and minima. Graph.
{{{f(x)=(x^2-1)/(x^2+1)}}}
Derivative is {{{4x/(x^2+1
Algebra ->
Polynomials-and-rational-expressions
-> SOLUTION: Find the intervals of increase and decrease. Use the first derivatives test to find the local maxima and minima. Graph.
{{{f(x)=(x^2-1)/(x^2+1)}}}
Derivative is {{{4x/(x^2+1
Log On
Question 862385: Find the intervals of increase and decrease. Use the first derivatives test to find the local maxima and minima. Graph.
Derivative is I think Answer by solver91311(24713) (Show Source):
You need to use a combination of the quotient rule and the chain rule.
1st derivative is equal to zero only when x = 0, hence the only extremum is at x = 0.
Second derivative:
Second derivative has the value 4 (> 0) at x = 0, hence the extremum is a minimum.
The denominator of the original function has no real zeros, hence there are no vertical asymptotes to the graph of this function.
I'll leave the graphing to you, but it sort of looks like a side view of an "innie" belly button with a minimum at (0,-1) and asymptotic to y = 1 as x increases or decreases without bound.
John
My calculator said it, I believe it, that settles it
