Approximate the real zeros f(x)=3x^4+x^2-1 to the nearest tenth

Step-by-step explanation:

Let us do by trial and error method

f(0) = -1 and f(1) = 3

f(-1) = 3

i.e. f(1) = f(-1)

Since the function is a polynomial it is continuous.

This is an even function because f(x) = f(-x)

So if we find one root, we can easily guess the other root as -value

There is a root between 0 and 1 and also -1 and 0.

f(0.6) and f(0.7) have change of sign.  SO root lies between 0,6 and 0.7

By filtering we find that the root is approximately 0,659.

BY symmetry about y axis, -0.659 is the other root

(x+0.659) and (x-0.659) are factors

When we divide by this product as x^2-0.434 we get quotient as

x^2+0,768

So the other two roots are imaginary.

Only real roots are

0,659 and -0,659

18

step-by-step explanation:

draw the shape and you will see that f and k are actually at adjacent vertices (points of the hexagon that are neighboring).   so find the distance between (1,2) and (4,2).   that is easy since (1,2) and (4,2) is just a vertical distance to count, which is 3.   it is a regular polygon which means all sides are congruent.   you have 6 sides (hexa) so multiply 6 by 3 to receive 18.