Find the number of possible negative real zeros for f(x)=6+x^4+2x^2-5x^3-12.

Answer: 0

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

Answer: no real roots

The first function takes the value zero at approximately X= -0.9 and again at around 4.63, with a minimum value between them somewhere around 3.46; also has a gradient of zero at X=0.

Are you sure the equation is correctly written down, given that it's apparently got two constants in it but no linear term?

Oops---left the "x" off of 12

Should read: f(x)=6+x^4+2x^2-5x^3-12x

Sorry

I suspected as much :) Your first function appears to have two zeros, but both are for positive X (approx. 0.5 and 5.0). The second one appears to have a couple of zeros somewhere around -1.4 and +1.4: would you like to check that out? (That's just from a quick sketch of the graph between X = -2 and X = +2.)

To determine the number of possible negative real zeros for a given function, we can use Descartes' Rule of Signs. Here's how you can apply it to f(x)=6+x^4+2x^2-5x^3-12:

1. Count the number of sign changes in the sequence of coefficients. In this case, we have two sign changes: from 6 to -5x^3 and from -5x^3 to -12.

2. Subtract an even number from the count of sign changes in step 1. Since there are two sign changes, subtracting an even number (0) gives us 2.

Therefore, according to Descartes' Rule of Signs, f(x) has either 2 or 0 negative real zeros. To determine the exact number, we need to find the number of positive real zeros. However, since f(x) does not have any positive terms, it follows that there are no negative real zeros possible for f(x). Hence, the answer is 0.

For the second question, to approximate the real zeros of f(x)=2x^4-3x^2-2 to the nearest tenth, we can use a numerical method such as the Newton-Raphson method or the bisection method. However, before applying any numerical method, we can check if there are any rational roots by using the rational root theorem.

To apply the rational root theorem:
1. Determine the possible rational roots. They are the factors of the constant term (in this case, -2) divided by the factors of the leading coefficient (2). The possible rational roots are -2, -1, 1, and 2.

Now, we can check these possible rational roots to see if any of them are actual roots of the equation:

f(-2) = 2(-2)^4 - 3(-2)^2 - 2
= 32 - 12 - 2
= 18

f(-1) = 2(-1)^4 - 3(-1)^2 - 2
= 2 - 3 - 2
= -3

f(1) = 2(1)^4 - 3(1)^2 - 2
= 2 - 3 - 2
= -3

f(2) = 2(2)^4 - 3(2)^2 - 2
= 32 - 12 - 2
= 18

Since none of the possible rational roots evaluated to give f(x) = 0, there are no rational roots (and hence no real roots) for the equation f(x)=2x^4-3x^2-2. Therefore, the answer is "no real roots."