This discussion focuses on identifying local extrema and saddle points for the function defined as f(x) = y, where the first derivative is incorrectly calculated as dy/dx = -2x^2 / (x^4 - 2x^2 + 1). The correct approach involves applying the quotient rule for differentiation. The asymptotes are identified at +/- 1, with the function increasing for x >= 2 and decreasing for x <= -2. The critical point at x = 0 requires further analysis to determine if it is a local maximum, minimum, or a saddle point.
PREREQUISITESStudents studying calculus, mathematics educators, and anyone interested in understanding local extrema and saddle points in functions.
kukumaluboy said:Homework Statement
Homework Equations
The Attempt at a Solution
1)
I found the asymptote as (+/- 1)
2)
Let f(x) = y;
dy/dx = -2x^2 / (x^4 - 2x^2 + 1) = 0
-2x^2 - 0
x = 0;
Since f() != 1, f(2) > 0 Increasing
Since f() != -1, f(-2) < 0 Decreasing
So i guess range is increasing or x >=2, decreasing for x<=-2
3)
Since x = 0, how find local max or min or what the toot is saddle point..
I think you've differentiated that incorrectly, remember to use the product (or quotient) rule.kukumaluboy said:Homework Statement
Homework Equations
The Attempt at a Solution
1)
I found the asymptote as (+/- 1)
2)
Let f(x) = y;
dy/dx = -2x^2 / (x^4 - 2x^2 + 1) = 0
-2x^2 - 0
x = 0;
Since f() != 1, f(2) > 0 Increasing
Since f() != -1, f(-2) < 0 Decreasing
So i guess range is increasing or x >=2, decreasing for x<=-2
3)
Since x = 0, how find local max or min or what the toot is saddle point..