Discuss the end behavior of the function: 3x^4 - 2x^2 + 5?

Question

I'm not sure what to say about that?

Answer 1

I assume you are in calculus. First the function has an extrema when

df/dx = 0 = 12x^3-4x --> 3x^2-1 = 0 so x = +/- 1/sqrt(3)

Now test for min/max by evaluating second derivative at each value of x from above

d^2f/dx^2 = 36x^2 - 4 --> this is always positive for both values of x that make first derivative zero, hence both extremas are local mins.

So there must be inflection points in between.

d^2f/dx^2 = 0 = 36x^2 - 4 = 9x^2 -1 --> x=+/- 1/3

Now at x = 0, f(0) = 5. Root of the function occurs at

3x^4-2x^2 = -5

x^2(3x^2-2) = -5

or let z= x^2, then

3z^2-2z+5 = 0 ---> (3z-5)(z+1)=0 ---> z = +5/3,-1

so x=+/-sqrt(5/3),+/-sqrt(-1)=+/-i

Imaginary roots indicate that the function only crosses the x-axis twice.

Answer 2

As x->±∞, 3x^4 - 2x^2 + 5 ->∞
End behavior: ↖ and ↗

Answer 3

First do 3*3*4-2*2*2+5= 33.How by 3*3 is for the^ sign.Ok?

Answer 4

y = 3x^4 - 2x^2 + 5
root is 0.9298942582 . . . . when y = 0
from the rule of sign . . . . . the curve will cross the x axis twice