# Functions

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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 -

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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