Determine the maximum possible number of turning points for the graph of the fucntion.
g(x)=-1/5x + 2
I got 0
How do I graph f(x)=x^5-4x^3-12x
Functions - Reiny, Tuesday, October 23, 2007 at 11:09pm
your first one is right
=x(x^4 - 4x^2 - 12)
if this had been the equation
x^5-4x^3-12x = 0
there would be solutions of
x=0, x=±√6 = ±2.45, and 2 complex roots
so the graph crosses at -2.45, 0, 2,45
it rises into the first quadrant, and drops into the third quadrants.
The two complex roots cause a "wiggle" in the first quadrant above the x-axis