Posted by **Em** on Tuesday, October 23, 2007 at 10:55pm.

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

for f(x)=x^5-4x^3-12x

=x(x^4 - 4x^2 - 12)

=x(x^2-6)(x^2+2)

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

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